Chemistry · Study Guide

Exam 1 Study Guide

This guide covers the first three chapters. You will measure matter, count its particles, and name what you find. Work through one chapter at a time. Read the "what to know" list, copy the equations out by hand, then try the worked examples with the answers covered. Finish each chapter with its ten self-test items. The cram sheet at the end holds every equation and constant in one place.

Chapter 1 Chapter 2 Chapter 3 Cram sheet

What this guide covers

Chapter 1 · matter, states, properties, SI units, significant figures, unit chains, density and temperature.
Chapter 2 · subatomic particles, isotopes, average atomic mass, the periodic table, ions, formulas and names.
Chapter 3 · the mole, molar mass, percent composition, empirical and molecular formulas, molarity and dilution.

Chapter 1Essential Ideas

What to know

Matter has mass and takes up space. It reaches you as a solid, a liquid or a gas. In a solid the particles sit in fixed spots. In a liquid they touch but slide past each other. In a gas they fly apart and fill the container.

Sort any sample in two steps. First ask if it is one pure substance or a mixture. A pure substance is either an element or a compound. A compound has atoms of more than one element bonded in a fixed ratio. A mixture has no fixed ratio. If you cannot see the parts, call it homogeneous. If you can, call it heterogeneous.

A physical change leaves the substance itself alone. Ice melting is a physical change, because it is still water. A chemical change builds new substances. Wood burning is a chemical change. Watch for clues like a new gas, a new color, a new solid or a flow of heat.

Matter is conserved. In a sealed container the products weigh what the reactants weighed. Nothing vanishes.

Every measurement needs a unit. Mass comes in kilograms, length in meters, time in seconds, amount in moles and temperature in kelvin. Volume and density are built from those. Prefixes scale a unit up or down. Kilo means 1,000 times. Centi means one hundredth. Milli means one thousandth. Micro means one millionth.

Significant figures record how sure you are. All nonzero digits count. Zeros between digits count. Leading zeros never count. Trailing zeros count only when a decimal point is shown. So 0.00450 has three significant figures, and 4500 has two unless it is written as 4500.

Rounding follows the weakest value. For multiplying and dividing, match the fewest significant figures. For adding and subtracting, match the fewest decimal places. Round once, at the end.

Accuracy and precision are not the same. Accuracy is how close you land to the true value. Precision is how close repeated tries land to each other. A tool with a steady offset is precise but not accurate.

Lock these in

Element and compound are pure. Everything else is a mixture.
New substance means chemical change. Same substance means physical change.
Trailing zeros count only with a decimal point in the number.
Multiply and divide by significant figures. Add and subtract by decimal places.
Density is an intensive property, so sample size never changes it.
Kelvin starts at absolute zero, which is −273.15 °C.

Equations, units and symbols

Equation	What each symbol means	Units
d = m ÷ V	d is density, m is mass, V is volume	d in g/mL or g/cm³; m in g; V in mL or cm³
m = d × V	same symbols, solved for mass	g = (g/mL) × mL
V = m ÷ d	same symbols, solved for volume	mL = g ÷ (g/mL)
K = °C + 273.15	K is the kelvin reading, °C is the Celsius reading	K and °C; the step size is the same in both
°F = (9/5 × °C) + 32	°F is the Fahrenheit reading	°F and °C
°C = 5/9 × (°F − 32)	same equation, solved for Celsius	°C and °F
1 cm³ = 1 mL	a volume identity worth memorizing	cm³ and mL
wanted = given × (wanted ÷ given)	dimensional analysis; the bracket is a conversion factor	the given unit cancels, the wanted unit survives

Worked examples

Worked example 1 · density of a metal bar

A bar of metal has a mass of 56.7 g. Its volume is 21.0 cm³. Find the density.

Step 1 · pick the equationYou hold mass and volume and want density, so use d = m ÷ V.

Step 2 · divide56.7 g ÷ 21.0 cm³ = 2.70 g/cm³.

Step 3 · roundBoth values hold three significant figures. So the answer keeps three: 2.70 g/cm³.

Check it a second wayMultiply back. 2.70 g/cm³ × 21.0 cm³ = 56.7 g. That is the mass you started with. The value also matches aluminum.

Worked example 2 · two rounding rules in one problem

A rectangle measures 12.5 cm by 3.46 cm. Find its area. Then add three masses: 24.681 g, 0.35 g and 1.4 g.

Step 1 · the area12.5 × 3.46 = 43.25 cm².

Step 2 · round the areaThis is multiplying. The weaker value is 12.5, with three significant figures. So report 43.3 cm².

Step 3 · the sum24.681 + 0.35 + 1.4 = 26.431 g.

Step 4 · round the sumThis is adding, so decimal places rule. The 1.4 g stops at the tenths place. So report 26.4 g.

Why they differMultiplying counts digits. Adding counts places. Using the wrong rule is the most common slip on this topic.

Worked example 3 · body temperature in all three scales

A thermometer reads 98.6 °F. Convert it to Celsius, then to kelvin.

Step 1 · subtract 3298.6 − 32 = 66.6.

Step 2 · scale by 5/95/9 × 66.6 = 37.0, so the reading is 37.0 °C.

Step 3 · add 273.1537.0 + 273.15 = 310.15.

Step 4 · round the sumAdding follows decimal places. The 37.0 stops at the tenths place. So report 310.2 K.

Check it a second wayRun it backward. (9/5 × 37.0) + 32 = 66.6 + 32 = 98.6 °F. The two agree.

Traps people fall into

Mixing the two rounding rules. Digits for multiplying. Places for adding.
Counting trailing zeros that do not count. 4500 has two figures. 4500. has four.
Using 273 instead of 273.15. The extra digits matter when your data is precise.
Rounding in the middle. Carry extra digits and round only the final answer.
Swapping accuracy and precision. Tight and wrong is precise, not accurate.
Thinking density grows with sample size. Mass and volume both grow, so the ratio holds.
Flipping a conversion factor. Write the units and cancel them. If they do not cancel, flip the factor.

Chapter 1 self-test · ten items

1. Is brass an element, a compound or a mixture?

A homogeneous mixture. Brass is copper and zinc melted together. The ratio can be changed, so it is not a compound. You cannot see the two metals apart, so it is homogeneous.

2. Sort these as physical or chemical: sugar dissolving in water, and iron rusting.

Dissolving is physical. Rusting is chemical. The sugar is still sugar, and boiling the water back off would return it. Rust is iron oxide, a brand new substance.

3. How many significant figures are in 0.004070?

Four. The two leading zeros only place the decimal point. The 4, the captive 0, the 7 and the trailing 0 all count, because a decimal point is shown.

4. Multiply 45.00 by 0.0030 and report the answer properly.

0.14. The raw product is 45.00 × 0.0030 = 0.135. The weaker value is 0.0030, which carries two significant figures. Rounding 0.135 to two figures gives 0.14.

5. Subtract 4.27 g from 108.5 g and report the answer properly.

104.2 g. The raw difference is 108.5 − 4.27 = 104.23 g. Subtracting follows decimal places. The 108.5 g stops at the tenths place, so the answer stops there too.

6. Convert 0.0450 km into centimeters.

4.50 × 10³ cm. Chain two factors. 0.0450 km × (1000 m / 1 km) = 45.0 m. Then 45.0 m × (100 cm / 1 m) = 4500 cm. Both factors are exact, so keep the three figures from 0.0450. Write it as 4.50 × 10³ cm.

7. Ethanol has a density of 0.789 g/mL. What is the mass of 125 mL?

98.6 g. Use m = d × V. 0.789 g/mL × 125 mL = 98.625 g. Both values hold three significant figures, so report 98.6 g. The milliliters cancel and grams are left.

8. Mercury has a density of 13.53 g/mL. What volume does 50.0 g fill?

3.70 mL. Use V = m ÷ d. 50.0 g ÷ 13.53 g/mL = 3.6955 mL. Three figures go in, so three come out: 3.70 mL. Check it: 13.53 × 3.70 = 50.06 g, which rounds back to 50.0 g.

9. Convert −40 °F to Celsius, then convert 25 °C to kelvin.

−40 °C, and 298 K. First: 5/9 × (−40 − 32) = 5/9 × (−72) = −40 °C. That is the one point where the two scales agree. Second: 25 + 273.15 = 298.15, and 25 stops at the ones place, so report 298 K.

10. A rock has a mass of 65.4 g. Dropped into water, it lifts the level from 25.0 mL to 46.5 mL. Find its density.

3.04 g/mL. The rock's volume is the water it pushed aside: 46.5 − 25.0 = 21.5 mL. Then d = 65.4 g ÷ 21.5 mL = 3.0419 g/mL. Three figures survive, so report 3.04 g/mL. It is denser than water, so it sinks.

Chapter 2Atoms, Molecules and Ions

What to know

Three particles build every atom. Protons carry a 1+ charge and sit in the nucleus. Neutrons carry no charge and sit there too. Electrons carry a 1− charge and move in the space outside. A proton and a neutron each weigh about 1 u. An electron weighs about 1/1836 as much, so the nucleus holds nearly all the mass.

The atomic number Z counts the protons. That count names the element and never changes. The mass number A counts protons plus neutrons. So neutrons = A − Z. Isotopes are atoms of one element with different neutron counts. They react the same way, because they hold the same electrons.

The mass on the periodic table is a weighted average. Multiply each isotope mass by its share of the sample, then add the products. A share must go in as a decimal, not a percent.

The table is a map of behavior. Columns are groups and share chemical habits. Rows are periods. Group 1 holds the alkali metals, group 2 the alkaline earth metals, group 17 the halogens and group 18 the noble gases. A stair step near the right side splits metals from nonmetals, and the elements along it are metalloids.

An ion is an atom that gained or lost electrons. Lose electrons and you get a positive cation. Gain them and you get a negative anion. Metals usually lose, and nonmetals usually gain. The charge is protons minus electrons.

An ionic formula is a charge balance. Find the smallest whole-number counts that make the total charge zero. A polyatomic ion is a bonded group with one charge of its own, like nitrate NO₃⁻ or sulfate SO₄²⁻. Wrap it in parentheses when you need more than one.

Naming splits three ways. Ask first whether the formula is an acid, an ionic compound or a molecular one. Ionic names give the metal, then the nonmetal with an -ide ending. A metal with more than one possible charge takes a Roman numeral, as in copper(II) chloride. Molecular names use Greek prefixes, and the first element drops "mono-". Acids that start with H and water follow their own set.

Lock these in

Z names the element. A − Z gives the neutron count.
Charge equals protons minus electrons.
Average atomic mass uses decimal fractions, so 75.76% goes in as 0.7576.
Ionic formulas balance charge to zero, using the smallest whole numbers.
Roman numerals mark the metal's charge, not the subscript.
Greek prefixes belong to molecular names only, never to ionic ones.

Equations, units and symbols

Equation	What each symbol means	Units
A = Z + N	A is mass number, Z is atomic number, N is the neutron count	all three are plain counts, so no unit
N = A − Z	same symbols, solved for neutrons	a plain count
charge = protons − electrons	the net charge on an atom or ion	a whole number of elementary charges
average atomic mass = Σ (f_i × m_i)	f_i is isotope i's decimal fraction, m_i is its mass	f has no unit; m and the answer are in u (or g/mol)
f = percent ÷ 100	turns a percent abundance into a fraction	f has no unit
total charge of an ionic formula = 0	the rule that sets every ionic subscript	charges must cancel exactly

Worked examples

Worked example 1 · average atomic mass of chlorine

Chlorine is 75.76% chlorine-35 at 34.969 u. The rest is 24.24% chlorine-37 at 36.966 u. Find the average atomic mass.

Step 1 · turn percents into fractions75.76% becomes 0.7576. 24.24% becomes 0.2424. The two add to 1.0000, which is a good sign.

Step 2 · weigh the light isotope0.7576 × 34.969 u = 26.4925 u.

Step 3 · weigh the heavy isotope0.2424 × 36.966 u = 8.9606 u.

Step 4 · add them26.4925 + 8.9606 = 35.4531 u, so the answer is 35.45 u.

Check it a second wayThe answer must land between 34.969 and 36.966. It does. It sits much nearer the light isotope, because that isotope owns three quarters of the sample. The table lists 35.45 for chlorine, so the work holds.

Worked example 2 · counting particles in an ion

Count the protons, neutrons and electrons in ⁵⁶Fe³⁺.

Step 1 · read Z from the tableIron is element 26, so there are 26 protons. That never changes.

Step 2 · subtract for neutronsN = A − Z = 56 − 26 = 30 neutrons.

Step 3 · use the chargeA 3+ charge means three electrons are missing. So electrons = 26 − 3 = 23 electrons.

Check it a second wayProtons minus electrons should give the charge. 26 − 23 = 3, and the ion is 3+. The two agree.

Worked example 3 · formulas and names, both directions

Write the formula for aluminum oxide. Then name Fe₂(SO₄)₃, SF₆ and HNO₃(aq).

Step 1 · the two ionsAluminum is in group 13, so it forms Al³⁺. Oxide is O²⁻.

Step 2 · balance the chargeThe least common multiple of 3 and 2 is 6. Two Al³⁺ give +6, and three O²⁻ give −6. So the formula is Al₂O₃.

Step 3 · name the iron saltThree sulfates give −6, so two iron ions must give +6. Each iron is 3+. The name is iron(III) sulfate.

Step 4 · name the molecular oneSF₆ is two nonmetals, so use prefixes. Sulfur drops "mono-" and six fluorines take "hexa-". The name is sulfur hexafluoride.

Step 5 · name the acidHNO₃ is H plus nitrate. An -ate ion becomes an -ic acid, so this is nitric acid.

Traps people fall into

Putting percents straight into the average. Divide by 100 first, or your answer comes out 100 times too big.
Mixing mass number with atomic mass. Mass number is a whole count of particles. Atomic mass is a measured mass in u.
Changing a subscript inside a polyatomic ion. Wrap the ion in parentheses and put the number outside.
Forgetting the charge when counting electrons. A cation is short of electrons. An anion has extra ones.
Using Greek prefixes on ionic names. CaCl₂ is calcium chloride, never calcium dichloride.
Leaving off the Roman numeral. Iron, copper, lead, tin and chromium all need one.
Not reducing the ratio. Mg²⁺ with O²⁻ is MgO, not Mg₂O₂.

Chapter 2 self-test · ten items

1. Count the protons, neutrons and electrons in ³¹P³⁻.

15 protons, 16 neutrons, 18 electrons. Phosphorus is element 15, so Z = 15. Then N = 31 − 15 = 16. The 3− charge means three extra electrons, so 15 + 3 = 18.

2. Boron is 19.9% boron-10 at 10.0129 u and 80.1% boron-11 at 11.0093 u. Find the average atomic mass.

10.81 u. First: 0.199 × 10.0129 = 1.9926 u. Second: 0.801 × 11.0093 = 8.8184 u. Add them: 1.9926 + 8.8184 = 10.8110 u, so 10.81 u. It sits near 11, because boron-11 owns four fifths of the sample.

3. Write the formula for magnesium nitride.

Mg₃N₂. Magnesium is group 2, so Mg²⁺. Nitride is N³⁻. The least common multiple of 2 and 3 is 6. Three Mg give +6 and two N give −6.

4. Name Ca(OH)₂ and write the formula for ammonium sulfate.

Calcium hydroxide, and (NH₄)₂SO₄. Calcium has only one charge, so no Roman numeral is needed. For the second one, ammonium is NH₄⁺ and sulfate is SO₄²⁻, so two ammonium ions are needed.

5. Name SO₃ and name N₂O₄.

Sulfur trioxide, and dinitrogen tetroxide. Both are two nonmetals, so prefixes apply. The first element drops "mono-", which is why it is sulfur and not monosulfur. The "a" in tetra drops before the vowel in oxide.

6. Name HClO₃(aq) and HCl(aq).

Chloric acid, and hydrochloric acid. ClO₃⁻ is chlorate, and an -ate ion gives an -ic acid. HCl holds no oxygen, so it takes the hydro- form with an -ic ending.

7. An ion holds 16 protons and 18 electrons. Name it and write its symbol.

Sulfide, S²⁻. Charge equals protons minus electrons: 16 − 18 = −2. Element 16 is sulfur, and a lone anion takes the -ide ending.

8. Which pair are isotopes of each other: ¹⁴C and ¹⁴N, or ¹²C and ¹³C?

Carbon-12 and carbon-13. Isotopes share the same Z and differ in neutrons. Carbon-14 and nitrogen-14 share a mass number but hold 6 and 7 protons, so they are different elements.

9. Element 20 sits in which group and period, and what charge does its ion carry?

Calcium, group 2, period 4, and it forms Ca²⁺. Group 2 metals lose both outer electrons. That is why the alkaline earth metals all form 2+ ions.

10. How many of each atom are in Al₂(CO₃)₃?

2 Al, 3 C and 9 O. The 3 outside the parentheses multiplies everything inside. So the one carbon becomes 3, and the three oxygens become 3 × 3 = 9. Aluminum sits outside, so its 2 stands as written.

Chapter 3Composition of Substances and Solutions

What to know

A mole is a counting word, like a dozen. One mole is 6.022 × 10²³ particles. That number is Avogadro's number, written N_A. Atoms are far too small to count one by one, so chemists weigh them instead.

Molar mass is the mass of one mole in grams. Add up the atomic masses in the formula and attach g/mol. Molar mass is the bridge between grams and moles. Multiply moles by it to get grams. Divide grams by it to get moles.

Every problem in this chapter runs through moles in the middle. Grams go to moles, moles go to particles, and moles go to other moles. Avogadro's number links moles to a count. Molar mass links moles to a mass.

Percent composition tells you what share of a compound's mass each element owns. Take the mass of that element in one mole, divide by the molar mass, then multiply by 100. The answer belongs to the compound itself, so sample size never enters it.

An empirical formula is the smallest whole-number ratio of atoms. Start from masses or percents. Turn each one into moles, then divide every result by the smallest. If a ratio lands near 1.5 or 1.33, multiply all of them by 2 or 3 to clear it.

A molecular formula tells you the real count in one molecule. Divide the true molar mass by the empirical formula mass to get n. Then multiply every subscript by n.

Molarity measures how crowded a solution is. It is moles of solute per liter of solution. The volume is the final volume in the flask, not the volume of water you poured in. Rearrange it three ways: find M, find moles, or find the volume.

Dilution adds solvent and keeps the solute. The moles stay the same, so M₁V₁ = M₂V₂. The volume units just have to match on both sides, so milliliters work fine here.

Lock these in

N_A = 6.022 × 10²³ particles per mole.
Moles sit in the middle of every conversion in this chapter.
Percent composition never depends on how much sample you have.
Empirical formulas need moles, so convert mass first.
Molarity uses liters of solution, measured to the mark.
In M₁V₁ = M₂V₂, the units on the two volumes only need to match each other.

Equations, units and symbols

Equation	What each symbol means	Units
n = m ÷ M	n is moles, m is mass, M is molar mass	mol = g ÷ (g/mol)
m = n × M	same symbols, solved for mass	g = mol × (g/mol)
N = n × N_A	N is the particle count, N_A is Avogadro's number	particles = mol × (6.022 × 10²³/mol)
mass % of X = (mass of X in one mole ÷ M) × 100%	X is one element in the formula, M is the molar mass	both masses in g; the answer is a percent
n_scale = M ÷ (empirical formula mass)	n_scale is the whole number that scales the formula up	both masses in g/mol; n_scale has no unit
M_c = moles of solute ÷ liters of solution	M_c is molarity, the concentration	mol/L, which is also written M
moles = M_c × V	same symbols, solved for moles	mol = (mol/L) × L
M₁V₁ = M₂V₂	1 is the stock, 2 is the diluted solution	M in mol/L; V in any one volume unit used on both sides

Worked examples

Worked example 1 · molar mass, then grams to moles

Find the molar mass of calcium nitrate, Ca(NO₃)₂. Then find the moles in 125 g.

Step 1 · count the atomsThe 2 outside multiplies both the N and the three O inside. So you have 1 Ca, 2 N and 6 O.

Step 2 · weigh each elementCa: 1 × 40.08 = 40.08. N: 2 × 14.007 = 28.014. O: 6 × 15.999 = 95.994.

Step 3 · add them40.08 + 28.014 + 95.994 = 164.088, so 164.09 g/mol.

Step 4 · grams to molesn = 125 g ÷ 164.09 g/mol = 0.7618 mol, which rounds to 0.762 mol.

Check it a second wayMultiply back: 0.762 mol × 164.09 g/mol = 125.0 g. Grams cancel on the way down and reappear on the way up.

Worked example 2 · percent composition of carbon dioxide

Find the mass percent of carbon and of oxygen in CO₂.

Step 1 · molar massC: 12.011. O: 2 × 15.999 = 31.998. Sum = 44.009, so 44.01 g/mol.

Step 2 · carbon's share(12.011 ÷ 44.009) × 100 = 27.29% C.

Step 3 · oxygen's share(31.998 ÷ 44.009) × 100 = 72.71% O.

Step 4 · check the sum27.29 + 72.71 = 100.00%. The percents must total 100, so this confirms both numbers.

Use itA 50.0 g tank of CO₂ holds 0.2729 × 50.0 = 13.6 g of carbon. The percent came from one mole, but it works on any amount.

Worked example 3 · percents to a molecular formula

A compound is 40.00% C, 6.71% H and 53.28% O. Its molar mass is 180.16 g/mol. Find both formulas.

Step 1 · assume 100.00 gThat hands you 40.00 g C, 6.71 g H and 53.28 g O.

Step 2 · grams to molesC: 40.00 ÷ 12.011 = 3.330 mol. H: 6.71 ÷ 1.008 = 6.657 mol. O: 53.28 ÷ 15.999 = 3.330 mol.

Step 3 · divide by the smallestThe smallest is 3.330. C gives 1.000. H gives 1.999. O gives 1.000.

Step 4 · read the ratioThat is 1 : 2 : 1, so the empirical formula is CH₂O.

Step 5 · empirical formula mass12.011 + 2(1.008) + 15.999 = 30.026 g/mol.

Step 6 · find the scalen = 180.16 ÷ 30.026 = 6.000. Multiply every subscript by 6 to get C₆H₁₂O₆.

Worked example 4 · make a solution, then dilute one

First, how much NaCl do you weigh for 500.0 mL of 0.250 M? Second, how much 6.00 M HCl do you pour to make 250.0 mL of 0.500 M?

Step 1 · moles neededmoles = 0.250 mol/L × 0.5000 L = 0.125 mol.

Step 2 · moles to gramsNaCl is 22.990 + 35.45 = 58.44 g/mol. So 0.125 mol × 58.44 g/mol = 7.305 g, or 7.31 g.

Step 3 · set up the dilutionM₁ = 6.00 M, M₂ = 0.500 M, V₂ = 250.0 mL. Rearrange to V₁ = M₂V₂ ÷ M₁.

Step 4 · solveV₁ = (0.500 × 250.0) ÷ 6.00 = 125 ÷ 6.00 = 20.83 mL, so 20.8 mL.

Check it a second wayStock side: 6.00 mol/L × 0.0208 L = 0.125 mol. Final side: 0.500 mol/L × 0.2500 L = 0.125 mol. The moles match, as dilution requires.

Traps people fall into

Using liters of solvent. Molarity needs liters of the finished solution, topped up to the mark.
Leaving volume in milliliters for molarity. Divide by 1000 first. In M₁V₁ = M₂V₂ you can keep milliliters, because the unit cancels.
Comparing masses for an empirical formula. Convert to moles first. Atoms differ in weight, so grams do not track counts.
Rounding a 1.5 ratio to 2. A clean half is real data. Multiply every ratio by 2 instead.
Missing a subscript outside parentheses. In Ca(NO₃)₂ the 2 multiplies six oxygens, not three.
Mixing up the two n values. One n is moles. The other is the whole number that scales a formula.
Forgetting that percent composition is fixed. A 1 g sample and a 1 kg sample have the same percents.

Chapter 3 self-test · ten items

1. Find the molar mass of ammonium carbonate, (NH₄)₂CO₃.

96.09 g/mol. The 2 outside gives 2 N and 8 H. N: 2 × 14.007 = 28.014. H: 8 × 1.008 = 8.064. C: 12.011. O: 3 × 15.999 = 47.997. Sum: 28.014 + 8.064 + 12.011 + 47.997 = 96.086 g/mol.

2. How many moles are in 45.0 g of water? How many molecules is that?

2.50 mol, and 1.51 × 10²⁴ molecules. Water is 2(1.008) + 15.999 = 18.015 g/mol. Then 45.0 ÷ 18.015 = 2.4979 mol, so 2.50 mol. Next: 2.50 mol × 6.022 × 10²³/mol = 1.5055 × 10²⁴, so 1.51 × 10²⁴.

3. What is the mass of 0.350 mol of NaOH?

14.0 g. NaOH is 22.990 + 15.999 + 1.008 = 40.00 g/mol. Then 0.350 mol × 40.00 g/mol = 14.00 g. Three figures went in, so report 14.0 g.

4. What percent of ammonia's mass is nitrogen?

82.24% N. NH₃ is 14.007 + 3(1.008) = 17.031 g/mol. Then (14.007 ÷ 17.031) × 100 = 82.24%. Hydrogen takes the other 17.76%, and the two add to 100.00%.

5. A compound is 85.6% C and 14.4% H. Find its empirical formula.

CH₂. Take 100 g. C: 85.6 ÷ 12.011 = 7.127 mol. H: 14.4 ÷ 1.008 = 14.286 mol. Divide both by 7.127: C gives 1.000 and H gives 2.004. That rounds cleanly to 1 : 2.

6. That same compound has a molar mass of 42.08 g/mol. Find the molecular formula.

C₃H₆. The empirical formula mass of CH₂ is 12.011 + 2(1.008) = 14.027 g/mol. Then n = 42.08 ÷ 14.027 = 3.000. Multiply both subscripts by 3.

7. What is the molarity of a solution holding 0.500 mol of solute in 2.00 L?

0.250 M. Divide moles by liters: 0.500 mol ÷ 2.00 L = 0.250 mol/L. Molarity counts crowding, not total amount, so a bigger flask at the same molarity holds more solute.

8. How much KMnO₄ do you weigh for 250.0 mL of 0.100 M solution?

3.95 g. Molar mass: 39.098 + 54.938 + 4(15.999) = 158.03 g/mol. Moles needed: 0.100 mol/L × 0.2500 L = 0.0250 mol. Mass: 0.0250 mol × 158.03 g/mol = 3.9508 g, so 3.95 g.

9. You dilute 25.0 mL of 2.00 M solution down to 0.400 M. What is the final volume?

125 mL. Use V₂ = M₁V₁ ÷ M₂ = (2.00 × 25.0) ÷ 0.400 = 50.0 ÷ 0.400 = 125 mL. Check it: the molarity fell by a factor of 5.00, so the volume must grow by 5.00, and 25.0 × 5.00 = 125.

10. How many moles of oxygen atoms are in 0.250 mol of Al₂(SO₄)₃?

3.00 mol of O atoms. The 3 outside multiplies four oxygens, giving 12 O per formula unit. Then 0.250 mol × 12 = 3.00 mol of oxygen atoms. Subscripts are mole ratios inside a formula, just as coefficients are between substances.

One pageCram sheet

Everything from chapters 1 to 3 in one place. Cover the right column and see if you can rebuild each row.

Equations

Name	Equation	Units
Density	d = m ÷ V	g/mL or g/cm³
Celsius to kelvin	K = °C + 273.15	K
Celsius to Fahrenheit	°F = (9/5 × °C) + 32	°F
Fahrenheit to Celsius	°C = 5/9 × (°F − 32)	°C
Neutron count	N = A − Z	a plain count
Ion charge	charge = protons − electrons	whole number
Average atomic mass	Σ (fraction × isotope mass)	u, or g/mol
Mass to moles	n = m ÷ M	mol
Moles to mass	m = n × M	g
Moles to particles	N = n × N_A	particles
Mass percent	(mass of X in one mole ÷ M) × 100%	percent
Formula scale factor	n = M ÷ empirical formula mass	no unit
Molarity	M_c = mol solute ÷ L solution	mol/L = M
Dilution	M₁V₁ = M₂V₂	matching volume units

Constants and conversions

Quantity	Value	Note
Avogadro's number	N_A = 6.022 × 10²³ /mol	particles in one mole
Absolute zero	0 K = −273.15 °C	so 0 °C = 273.15 K
Volume identity	1 cm³ = 1 mL	1 L = 1000 mL = 1 dm³
Density of water	1.00 g/mL near room temperature	the float-or-sink line
Prefixes	k = 10³, c = 10⁻², m = 10⁻³, µ = 10⁻⁶, n = 10⁻⁹	attach to any base unit
Mass unit	1 u = the mass of 1/12 of a carbon-12 atom	1 u per particle matches 1 g/mol per mole
Atomic masses used here	H 1.008, C 12.011, N 14.007, O 15.999, Na 22.990, Cl 35.45, K 39.098, Ca 40.08, Mn 54.938	from the periodic table

Rules that are not equations

Topic	The rule
Significant figures	Nonzero digits count. Captive zeros count. Leading zeros never count. Trailing zeros count only with a decimal point.
Rounding a product	Match the fewest significant figures in the data.
Rounding a sum	Match the fewest decimal places in the data.
Exact numbers	Counted items and defined conversions have unlimited figures, so they never limit an answer.
Unit chains	Write every factor as a fraction. If the given unit does not cancel, flip the factor.
Ionic formulas	Balance charge to zero with the smallest whole numbers, then reduce.
Ionic names	Metal first, then the nonmetal with -ide. Add a Roman numeral when the metal has more than one charge.
Molecular names	Greek prefixes on both elements, except the first element drops mono-.
Acid names	No oxygen: hydro- plus -ic. An -ate ion gives -ic acid. An -ite ion gives -ous acid.
Empirical formulas	Mass to moles, divide by the smallest, then clear any half or third by multiplying all of them.