Chemistry worksheet focusing on empirical and molecular formulas with four problems related to percent composition and empirical formula determination.

Let’s solve each problem step-by-step with proper units and significant figures as requested.

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1.) Find the percent composition of plumbic bromide.

“Plumbic” refers to lead(IV), so the formula is PbBr₄.

We need to find the mass percent of each element: Pb and Br.

Step 1: Molar mass of PbBr₄

- Atomic mass of Pb = 207.2 g/mol
- Atomic mass of Br = 79.90 g/mol × 4 = 319.6 g/mol

→ Molar mass of PbBr₄ = 207.2 + 319.6 = 526.8 g/mol

Step 2: Percent composition

- % Pb = (mass of Pb / molar mass) × 100 = (207.2 / 526.8) × 100
→ 207.2 ÷ 526.8 ≈ 0.3933 → 39.33% Pb

- % Br = (mass of Br / molar mass) × 100 = (319.6 / 526.8) × 100
→ 319.6 ÷ 526.8 ≈ 0.6067 → 60.67% Br

✔ Answer: Pb = 39.33%, Br = 60.67%

*(Sig figs: All given atomic masses have 4 sig figs, so we report to 4 sig figs.)*

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2.) Determine the percent composition of calcium nitrate.

Formula of calcium nitrate: Ca(NO₃)₂

Step 1: Molar mass

- Ca = 40.08 g/mol
- N = 14.01 g/mol × 2 = 28.02 g/mol
- O = 16.00 g/mol × 6 = 96.00 g/mol

→ Molar mass = 40.08 + 28.02 + 96.00 = 164.10 g/mol

Step 2: Percent composition

- % Ca = (40.08 / 164.10) × 100 ≈ 24.42%
- % N = (28.02 / 164.10) × 100 ≈ 17.07%
- % O = (96.00 / 164.10) × 100 ≈ 58.50%

✔ Answer: Ca = 24.42%, N = 17.07%, O = 58.50%

*(Sig figs: All inputs have 4 sig figs, so answers reported to 4 sig figs.)*

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3.) A compound was analyzed and found to contain 9.8 g of nitrogen, 0.70 g of hydrogen, and 33.6 g of oxygen. What is the empirical formula?

Step 1: Convert masses to moles

- Moles of N = 9.8 g / 14.01 g/mol ≈ 0.6995 mol
- Moles of H = 0.70 g / 1.008 g/mol ≈ 0.6944 mol
- Moles of O = 33.6 g / 16.00 g/mol = 2.100 mol

Step 2: Divide by smallest number of moles

Smallest mole value is ~0.6944 (H)

- N: 0.6995 / 0.6944 ≈ 1.007
- H: 0.6944 / 0.6944 = 1.000
- O: 2.100 / 0.6944 ≈ 3.024

These are very close to whole numbers: N : H : O ≈ 1 : 1 : 3

So empirical formula = NHO₃

But that’s not standard — let’s check if it should be written as HNO₃ (nitric acid), which has 1 H, 1 N, 3 O.

Yes! That matches.

✔ Empirical formula: HNO₃

*(Note: Since the ratios are so close to 1:1:3, and HNO₃ is a known compound, this is correct.)*

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4.) Determine the empirical formula of a compound containing 3.6 g carbon, 0.90 g hydrogen, and 2.4 g oxygen.

Step 1: Convert to moles

- Moles C = 3.6 g / 12.01 g/mol ≈ 0.2997 mol
- Moles H = 0.90 g / 1.008 g/mol ≈ 0.8929 mol
- Moles O = 2.4 g / 16.00 g/mol = 0.1500 mol

Step 2: Divide by smallest mole value

Smallest is 0.1500 (O)

- C: 0.2997 / 0.1500 ≈ 1.998 ≈ 2.0
- H: 0.8929 / 0.1500 ≈ 5.953 ≈ 6.0
- O: 0.1500 / 0.1500 = 1.0

→ Ratio C : H : O = 2 : 6 : 1

✔ Empirical formula: C₂H₆O

*(This is the empirical formula for ethanol or dimethyl ether — both have same empirical formula.)*

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## ✔ Final Answers:

1.) Plumbic bromide (PbBr₄):
 Pb = 39.33%, Br = 60.67%

2.) Calcium nitrate (Ca(NO₃)₂):
 Ca = 24.42%, N = 17.07%, O = 58.50%

3.) Compound with 9.8g N, 0.70g H, 33.6g O:
 Empirical formula = HNO₃

4.) Compound with 3.6g C, 0.90g H, 2.4g O:
 Empirical formula = C₂H₆O

All work shown with correct units and significant figures.