What is the empirical formula of a compound that is 18.7% Li, 16.3% C, and 65.0% O?

Here are the solutions to the problems:

Question 17:
A student works a problem and finds that a compound has 0.074 moles of carbon and 0.148 moles of oxygen. What is the empirical formula of the compound?

• Step 1: Write down the moles of each element.

  • Moles of Carbon (C) = $0.074 \text{ mol}$
  • Moles of Oxygen (O) = $0.148 \text{ mol}$

• Step 2: Divide by the smallest number of moles to find the mole ratio.

  • For C: $$\frac{0.074 \text{ mol}}{0.074 \text{ mol}} = 1$$
  • For O: $$\frac{0.148 \text{ mol}}{0.074 \text{ mol}} = 2$$

• Step 3: Write the empirical formula. The ratio of C:O is 1:2. The empirical formula is $\boxed{\text{CO}_2}$.

Question 18:
What is the empirical formula for a compound that is 18.7% Li, 16.3% C, and 65.0% O?

• Step 1: Assume a 100 g sample.

  • Mass of Li = $18.7 \text{ g}$
  • Mass of C = $16.3 \text{ g}$
  • Mass of O = $65.0 \text{ g}$

• Step 2: Convert mass to moles using atomic masses (Li $\approx 6.94 \text{ g/mol}$, C $\approx 12.01 \text{ g/mol}$, O $\approx 16.00 \text{ g/mol}$).

  • Moles of Li: $$\frac{18.7 \text{ g}}{6.94 \text{ g/mol}} \approx 2.69 \text{ mol}$$
  • Moles of C: $$\frac{16.3 \text{ g}}{12.01 \text{ g/mol}} \approx 1.36 \text{ mol}$$
  • Moles of O: $$\frac{65.0 \text{ g}}{16.00 \text{ g/mol}} \approx 4.06 \text{ mol}$$

• Step 3: Divide by the smallest number of moles (1.36 mol).

  • For Li: $$\frac{2.69}{1.36} \approx 1.98 \approx 2$$
  • For C: $$\frac{1.36}{1.36} = 1$$
  • For O: $$\frac{4.06}{1.36} \approx 2.98 \approx 3$$

• Step 4: Write the empirical formula. The ratio of Li:C:O is 2:1:3. The empirical formula is $\boxed{\text{Li}_2\text{CO}_3}$.

Question 19:
A 100.0-g sample of a compound is made up of 35.9 g of aluminum and 64.1 g of sulfur. The empirical formula of the compound is

• Step 1: Write down the mass of each element.
  • Mass of Al = $35.9 \text{ g}$
  • Mass of S = $