Mole conversion practice worksheet for honors chemistry students.

I'll solve each problem step by step and explain the solutions.

## Problem 1: Calculate the number of either molecules or formula units for each

a. 3.75 moles of glucose (C₆H₁₂O₆)

- Use Avogadro's number: 6.02 × 10²³ molecules/mol
- Calculation: 3.75 mol × 6.02 × 10²³ molecules/mol = 2.2575 × 10²⁴ molecules
- Answer: 2.26 × 10²⁴ molecules C₆H₁₂O₆

b. 1.89 × 10⁻⁴ moles of SO₂

- Calculation: 1.89 × 10⁻⁴ mol × 6.02 × 10²³ molecules/mol = 1.13778 × 10²⁰ molecules
- Answer: 1.14 × 10²⁰ molecules SO₂

c. 1.35 moles of water (H₂O)

- Calculation: 1.35 mol × 6.02 × 10²³ molecules/mol = 8.127 × 10²³ molecules
- Answer: 8.13 × 10²³ molecules H₂O

d. 7.37 grams of strontium chloride (SrCl₂)

- First, find molar mass of SrCl₂: Sr = 87.62 g/mol, Cl = 35.45 g/mol × 2 = 70.90 g/mol
- Molar mass = 87.62 + 70.90 = 158.52 g/mol
- Moles = 7.37 g ÷ 158.52 g/mol = 0.0465 mol
- Formula units = 0.0465 mol × 6.02 × 10²³ formula units/mol = 2.80 × 10²² formula units
- Answer: 2.80 × 10²² formula units SrCl₂

## Problem 2: Calculate the atoms of oxygen in each

a. 5.50 grams of Sodium chlorate (NaClO₃)

- Molar mass NaClO₃: Na = 23.0, Cl = 35.5, O = 16.0 × 3 = 48.0 → 106.5 g/mol
- Moles = 5.50 g ÷ 106.5 g/mol = 0.0516 mol
- Each molecule has 3 O atoms
- Oxygen atoms = 0.0516 mol × 6.02 × 10²³ molecules/mol × 3 = 9.32 × 10²² atoms
- Answer: 9.32 × 10²² atoms of O

b. 0.245 moles of Nickel(II) sulfate (NiSO₄)

- Each molecule has 4 O atoms
- Oxygen atoms = 0.245 mol × 6.02 × 10²³ molecules/mol × 4 = 5.89 × 10²³ atoms
- Answer: 5.89 × 10²³ atoms of O

c. 17.3 × 10²⁴ particles of iron(II) carbonate (FeCO₃)

- Each molecule has 3 O atoms
- Oxygen atoms = 17.3 × 10²⁴ particles × 3 = 5.19 × 10²⁵ atoms
- Answer: 5.19 × 10²⁵ atoms of O

d. 17.0 × 10²⁴ formula units of lead(IV) acetate (Pb(C₂H₃O₂)₄)

- Each formula unit has 8 O atoms (4 × 2 from acetate groups)
- Oxygen atoms = 17.0 × 10²⁴ × 8 = 1.36 × 10²⁶ atoms
- Answer: 1.36 × 10²⁶ atoms of O

## Problem 3: Calculate the moles of calcium in each

a. 5.43 grams of Calcium chloride (CaCl₂)

- Molar mass CaCl₂: Ca = 40.0, Cl = 35.5 × 2 = 71.0 → 111.0 g/mol
- Moles of CaCl₂ = 5.43 g ÷ 111.0 g/mol = 0.0489 mol
- Each mole CaCl₂ contains 1 mole Ca
- Answer: 0.0489 mol Ca

b. 3.64 × 10²⁴ atoms of Calcium

- Moles = 3.64 × 10²⁴ atoms ÷ 6.02 × 10²³ atoms/mol = 6.05 mol
- Answer: 6.05 mol Ca

c. 8.13 moles of Calcium bromide (CaBr₂)

- Each mole CaBr₂ contains 1 mole Ca
- Answer: 8.13 mol Ca

The key concepts used:
1. Avogadro's number (6.02 × 10²³ particles/mol)
2. Molar mass calculations
3. Stoichiometric relationships between compounds and their constituent elements
4. Converting between mass, moles, and number of particles