3ucky
Answered

A sample of an unknown substance contains 38.43 g of lead, 17.83 g of carbon, and 3.74 g of hydrogen. What is the empirical formula?

A. [tex]PbC _4 H _{10}[/tex]

B. [tex]Pb _2 C _8 H _{15}[/tex]

C. [tex]PbC _8 H _{20}[/tex]

D. [tex]Pb _2 C _4 H _{15}[/tex]



Answer :

To determine the empirical formula, the steps involve converting the masses of each element to moles, then finding the simplest whole number ratio of moles of each element.

### Step 1: Calculate the number of moles of each element

1. Lead (Pb)
Given mass = 38.43 g
Molar mass of Pb = 207.2 g/mol

[tex]\[ \text{Moles of Pb} = \frac{38.43 \text{ g}}{207.2 \text{ g/mol}} = 0.185472972972973 \text{ moles} \][/tex]

2. Carbon (C)
Given mass = 17.83 g
Molar mass of C = 12.01 g/mol

[tex]\[ \text{Moles of C} = \frac{17.83 \text{ g}}{12.01 \text{ g/mol}} = 1.484596169858451 \text{ moles} \][/tex]

3. Hydrogen (H)
Given mass = 3.74 g
Molar mass of H = 1.008 g/mol

[tex]\[ \text{Moles of H} = \frac{3.74 \text{ g}}{1.008 \text{ g/mol}} = 3.7103174603174605 \text{ moles} \][/tex]

### Step 2: Calculate the mole ratio

To find the simplest ratio, divide the number of moles of each element by the smallest number of moles calculated.

[tex]\[ \text{Smallest number of moles} = 0.185472972972973 \text{ moles (Pb)} \][/tex]

1. Ratio for Pb

[tex]\[ \text{Ratio for Pb} = \frac{0.185472972972973 \text{ moles}}{0.185472972972973} = 1.0 \][/tex]

2. Ratio for C

[tex]\[ \text{Ratio for C} = \frac{1.484596169858451 \text{ moles}}{0.185472972972973} = 8.004380077925346 \][/tex]

3. Ratio for H

[tex]\[ \text{Ratio for H} = \frac{3.7103174603174605 \text{ moles}}{0.185472972972973} = 20.004626015554976 \][/tex]

### Step 3: Round the ratios to the nearest whole numbers

The ratios are close to 1, 8, and 20. Therefore:

- 1 for Pb
- 8 for C
- 20 for H

### Conclusion: Empirical Formula

From the ratios, the empirical formula is:

[tex]\[ \text{Empirical formula} = Pb C_8 H_{20} \][/tex]

So, the correct choice from the given options is:

[tex]\[ \boxed{PbC_8H_{20}} \][/tex]