The calculations you actually need in first-semester chemistry
Start at the beginning or jump to your current chapter. Every topic explains the mathematics, models the units and algebra, and gives you a separate problem to solve. Nothing here assumes you passed a college algebra prerequisite.
Part 1 · Numbers, atoms, formulas, and reactions
Numbers & scientific notation
Moles and molecular quantities often require numbers larger or smaller than ordinary decimals.
(3.0 × 104)/(1.5 × 10−2) = (3.0/1.5) × 104−(−2) = 2.0 × 106.
Calculate (6.0 × 10−3)/(2.0 × 102).
Show worked answer
(6.0/2.0) × 10−3−2 = 3.0 × 10−5.
Watch out: Put the whole denominator in parentheses when entering scientific notation.
Dimensional analysis
Unit conversion is multiplication by ratios that equal 1. Choose each ratio so the original unit cancels.
0.0850 kg × (1000 g/1 kg) = 85.0 g.
Convert 125 mL to L.
Show worked answer
125 mL × (1 L/1000 mL) = 0.125 L.
Watch out: Units are part of the calculation; do not drop them during intermediate steps.
Significant figures & precision
For multiplication/division, match the fewest significant figures; for addition/subtraction, match the least precise decimal place.
2.50 × 3.0 = 7.5 (2 significant figures); 12.11 + 0.3 = 12.4 (tenths place).
Round 4.45/2.0 to the appropriate significant figures.
Show worked answer
4.45/2.0 = 2.225, reported as 2.2 (2 significant figures).
Watch out: Do not round intermediate quantities; exact conversion factors do not limit significant figures.
Rearranging equations
Solve a formula for the variable you need before substituting numbers.
M = n/V → multiply by V → n = MV; similarly, PV = nRT → V = nRT/P.
Solve q = mcΔT for ΔT.
Show worked answer
Divide both sides by mc: ΔT = q/(mc).
Watch out: Be careful with parentheses when the denominator has several factors.
Weighted average atomic mass
Multiply each isotope’s mass by its decimal fractional abundance and add.
A model element: (10.000 amu × 0.750) + (11.000 amu × 0.250) = 10.25 amu.
An element is 80.0% isotope X at 20.0 amu and 20.0% isotope Y at 22.0 amu. Calculate its average mass.
Show worked answer
(0.800 × 20.0) + (0.200 × 22.0) = 16.0 + 4.4 = 20.4 amu.
Watch out: Use fractional abundances (0.800), not percentage numbers (80.0), in the sum.
Molar mass
Add atomic masses; multiply by subscripts and parentheses before summing.
M[Ca(NO3)2] = 40.08 + 2(14.01 + 3×16.00) = 164.10 g/mol.
Find the molar mass of CO2 from C = 12.01 and O = 16.00.
Show worked answer
12.01 + 2(16.00) = 44.01 g/mol.
Watch out: A subscript outside parentheses multiplies every atom inside the group.
Mass ↔ moles ↔ particles
Use molar mass to connect grams and moles; use Avogadro’s constant to connect moles and entities.
9.00 g H2O × (1 mol/18.02 g) = 0.499 mol H2O.
How many molecules are in 0.500 mol of a molecular substance?
Show worked answer
0.500 mol × 6.022 × 1023 molecules/mol = 3.01 × 1023 molecules.
Watch out: Grams ↔ moles requires molar mass; particles ↔ moles requires Avogadro’s constant.
Percent composition
Divide a constituent element’s mass in one mole of compound by the total molar mass.
CO2: mass % C = (12.01/44.01)×100% = 27.3%.
Find mass % O in CO2 using the same masses.
Show worked answer
(32.00/44.01)×100% = 72.7%.
Watch out: The percentages of all elements must sum to approximately 100%.
Empirical vs molecular formula
Assume a 100 g sample for percent data, convert each element to moles, then divide by the smallest molar amount.
For 40.0% C, 6.7% H, 53.3% O: C ≈ 3.33 mol; H ≈ 6.63 mol; O ≈ 3.33 mol. Ratio ≈ 1:2:1 → CH2O.
An empirical formula is CH2O (≈30.03 g/mol) and molecular molar mass is ≈180.18 g/mol. What is the molecular formula?
Show worked answer
180.18/30.03 ≈ 6, so multiply every subscript by 6: C6H12O6.
Watch out: If a mole ratio is near 1.5 or 1.33, multiply all ratios by a small whole number to obtain integers.
Mole ratios from equations
The coefficients in a balanced chemical equation give mole ratios—not mass ratios.
2H2 + O2 → 2H2O. From 3.00 mol O2 with excess H2: 3.00 × (2/1) = 6.00 mol H2O.
For 2Al + 3Cl2 → 2AlCl3, how much AlCl3 can form from 1.50 mol Al with excess Cl2?
Show worked answer
1.50 mol Al × (2 mol AlCl3/2 mol Al) = 1.50 mol AlCl3.
Watch out: Always balance the chemical equation before doing stoichiometric math.
Limiting reactant
Find how much product each reactant could make independently; the smaller result controls the maximum yield.
With 3.00 mol H2 and 1.00 mol O2 for 2H2 + O2 → 2H2O: H2 allows 3.00 mol H2O; O2 allows 2.00 mol. O2 limits; 2.00 mol H2O forms.
How much H2 is left over in that reaction?
Show worked answer
1.00 mol O2 uses 2.00 mol H2; 3.00 − 2.00 = 1.00 mol H2 remains.
Watch out: Never select the limiting reactant solely by whichever starting mole amount is smaller.
Percent yield and percent error
Yield compares actual product with theoretical product. Percent error compares a measured value with a reference value.
Percent yield = (8.40 g actual/10.0 g theoretical)×100% = 84.0%.
A measured density is 1.08 g/mL; accepted density is 1.00 g/mL. Find percent error magnitude.
Show worked answer
|1.08 − 1.00|/1.00 ×100% = 8% (based on given reference precision; do not overinterpret the last digit).
Watch out: Keep actual/theoretical yield distinct from measured/reference percent error.
Part 2 · Solutions and gases
Molarity
Molarity describes moles of solute per liter of total solution.
M = n/V. For 0.250 mol dissolved to make 0.500 L: 0.250/0.500 = 0.500 mol/L = 0.500 M.
How many moles are present in 250. mL of a 0.200 M solution?
Show worked answer
n = MV = (0.200 mol/L)(0.250 L) = 0.0500 mol.
Watch out: Use solution volume, not the volume of solvent you poured in.
Dilution calculations
When only solvent is added, solute moles are conserved: M1V1 = M2V2.
(2.00 M)(25.0 mL) = (0.500 M)(V2) → V2 = 100. mL of final solution.
How many mL of 1.50 M stock are needed to prepare 300. mL of 0.250 M solution?
Show worked answer
V1 = (0.250×300.)/1.50 = 50.0 mL stock; dilute to 300. mL final volume.
Watch out: V2 is final total solution volume, not the water volume to add.
Temperature and ideal gas law
Convert °C to K, then select consistent pressure, volume and gas-constant units.
PV = nRT. At n = 1.00 mol, T = 298 K, P = 1.00 atm, R = 0.08206 L·atm·mol−1·K−1: V = nRT/P ≈ 24.5 L.
At 1.00 atm and 300. K, how much volume does 0.500 mol ideal gas occupy?
Show worked answer
V = (0.500)(0.08206)(300.)/(1.00) = 12.3 L.
Watch out: Use Kelvin—not degrees Celsius—in gas-law formulas.
Combined gas law and partial pressures
Gas variables can scale with one another; in a mixture, total pressure is the sum of individual partial pressures.
At constant n and P, V2 = V1T2/T1. 2.00 L × 350. K/300. K = 2.33 L.
A mixture has partial pressures 0.200, 0.300, and 0.500 atm. Find total pressure.
Show worked answer
Ptotal = 0.200 + 0.300 + 0.500 = 1.000 atm (three decimal places).
Watch out: A changing gas temperature must always be expressed as an absolute temperature.
Part 3 · Energy and light
Calorimetry
The heat absorbed or released by a sample depends on mass, specific heat, and temperature change.
q = mcΔT. For 50.0 g water, c = 4.184 J/(g·°C), ΔT = +5.00 °C: q = 50.0×4.184×5.00 = 1046 J = +1.05 kJ.
How much heat does 100. g of water absorb for a +2.00 °C change? Use c = 4.184 J/(g·°C).
Show worked answer
q = 100.×4.184×2.00 = 836.8 J → +837 J.
Watch out: The sign of ΔT determines the sign of q for the material being measured.
Hess’s law and energy arithmetic
Reaction enthalpies add when chemical equations are added. Reverse an equation → reverse its ΔH sign; multiply equation → multiply ΔH.
If ΔH1 = −100 kJ and ΔH2 = +40 kJ for steps that add to the target reaction, ΔHtarget = −60 kJ.
A listed reaction has ΔH = −125 kJ. What is ΔH for the exact reverse reaction?
Show worked answer
Reverse the sign: +125 kJ.
Watch out: Do not reverse just the sign of the number without reversing the reaction as well.
Pressure–volume work & first law
Expansion work against constant external pressure is w = −PextΔV. For the system, ΔE = q + w.
A gas expands by +2.00 L against 1.00 atm: w = −2.00 L·atm × 101.325 J/(L·atm) = −203 J.
If the gas above absorbs +350 J of heat, what is ΔE?
Show worked answer
ΔE = +350 + (−202.65) = +147.35 J ≈ +147 J.
Watch out: Check signs: expansion work on the system is negative; heat absorbed by the system is positive.
Light, wavelength and photon energy
Relate wavelength and frequency with c = λν, then calculate energy with E = hν = hc/λ.
For λ = 500 nm = 5.00×10−7 m, h = 6.626×10−34 J·s and c = 2.998×108 m/s, E ≈ 3.97×10−19 J per photon.
Calculate E for one photon at λ = 400 nm.
Show worked answer
E = hc/(4.00×10−7 m) ≈ 4.97×10−19 J.
Watch out: Convert nm to meters before substituting SI values of h and c.
How to practice for an exam
Choose five problems that match your next exam. First solve them with the formula and units visible. Then hide your notes and solve them again. If you miss an answer, identify whether the problem was algebra, unit conversion, chemical interpretation, or calculator entry. Spend your next review session on that specific skill.
Continue with General Chemistry II Math, or return to Chemistry Math Foundations.
Many colleges place these topics in slightly different chapters or semesters. Check your instructor’s coverage. Exact constants used here include 1 L·atm = 101.325 J; 1 nm = 10−9 m. Scientific notation is used to model meaningful precision rather than display every calculator digit.