Calculate the empirical formula of organic compound containing C= 61.6% and H = 12.8%

Question

ScanSolve AI Tutor · Accepted Answer

Here are the empirical formulas for the compounds:

1) A compound contains C = 20g and H = 4g.

Convert masses to moles.
$\text{Moles of C} = \frac{20\, \text{g}}{12.01\, \text{g/mol}} = 1.665\, \text{mol}$ $\text{Moles of H} = \frac{4\, \text{g}}{1.008\, \text{g/mol}} = 3.968\, \text{mol}$
Divide by the smallest number of moles to find the mole ratio.
$\text{Ratio of C} = \frac{1.665\, \text{mol}}{1.665\, \text{mol}} = 1$ $\text{Ratio of H} = \frac{3.968\, \text{mol}}{1.665\, \text{mol}} \approx 2.38$
Since 2.38 is close to 2.5, we multiply both ratios by 2 to get whole numbers.
$\text{C ratio} = 1 \times 2 = 2$ $\text{H ratio} = 2.38 \times 2 \approx 4.76 \approx 5$
Write the empirical formula.
The empirical formula is $\boxed{\text{C}_2\text{H}_5}$ .

2) An organic compound contains C = 61.6% and H = 12.8%.
Assume a 100 g sample.
The mass of C is 61.6 g and H is 12.8 g. The remaining percentage must be Oxygen.
$\text{Mass of O} = 100\% - 61.6\% - 12.8\% = 25.6\%$
So, in a 100 g sample, C = 61.6 g, H = 12.8 g, O = 25.6 g.
Convert masses to moles.
$\text{Moles of C} = \frac{61.6\, \text{g}}{12.01\, \text{g/mol}} = 5.129\, \text{mol}$ $\text{Moles of H} = \frac{12.8\, \text{g}}{1.008\, \text{g/mol}} = 12.698\, \text{mol}$ $\text{Moles of O} = \frac{25.6\, \text{g}}{16.00\, \text{g/mol}} = 1.600\, \text{mol}$
Divide by the smallest number of moles to find the mole ratio.
$\text{Ratio of C} = \frac{5.129\, \text{mol}}{1.600\, \text{mol}} \approx 3.20$ $\text{Ratio of H} = \frac{12.698\, \text{mol}}{1.600\, \text{mol}} \approx 7.94$ $\text{Ratio of O} = \frac{1.600\, \text{mol}}{1.600\, \text{mol}} = 1$
Since 3.20 is close to 3.25 (or $3 \frac{1}{4}$ ) and 7.94 is close to 8, we can multiply by 4 to get whole numbers.
$\text{C ratio} = 3.20 \times 4 = 12.8 \approx 13$ $\text{H ratio} = 7.94 \times 4 = 31.76 \approx 32$ $\text{O ratio} = 1 \times 4 = 4$
Write the empirical formula.
The empirical formula is $\boxed{\text{C}_{13}\text{H}_{32}\text{O}_4}$ .

3) A compound was found to contain C = 84.45% and N = 6.07%. (Assuming H is the remainder, as the given H percentage was too high).
Calculate the percentage of H.
$\text{Percentage of H} = 100\% - 84.45\% - 6.07\% = 9.48\%$
So, in a 100 g sample, C = 84.45 g, H = 9.48 g, N = 6.07 g.
Convert masses to moles.
$\text{Moles of C} = \frac{84.45\, \text{g}}{12.01\, \text{g/mol}} = 7.032\, \text{mol}$ $\text{Moles of H} = \frac{9.48\, \text{g}}{1.008\, \text{g/mol}} = 9.405\, \text{mol}$ $\text{Moles of N} = \frac{6.07\, \text{g}}{14.01\, \text{g/mol}} = 0.433\, \text{mol}$
Divide by the smallest number of moles to find the mole ratio.
$\text{Ratio of C} = \frac{7.032\, \text{mol}}{0.433\, \text{mol}} \approx 16.24$ $\text{Ratio of H} = \frac{9.405\, \text{mol}}{0.433\, \text{mol}} \approx 21.72$ $\text{Ratio of N} = \frac{0.433\, \text{mol}}{0.433\, \text{mol}} = 1$
These ratios are close to whole numbers (16, 22, 1).
Write the empirical formula.
The empirical formula is $\boxed{\text{C}_{16}\text{H}_{22}\text{N}}$ .

4) A compound contains C = 6.06g, H = 0.40g, and Oxygen = 6.73g.
Convert masses to moles.
$\text{Moles of C} = \frac{6.06\, \text{g}}{12.01\, \text{g/mol}} = 0.5046\, \text{mol}$ $\text{Moles of H} = \frac{0.40\, \text{g}}{1.008\, \text{g/mol}} = 0.3968\, \text{mol}$ $\text{Moles of O} = \frac{6.73\, \text{g}}{16.00\, \text{g/mol}} = 0.4206\, \text{mol}$
Divide by the smallest number of moles to find the mole ratio.
$\text{Ratio of C} = \frac{0.5046\, \text{mol}}{0.3968\, \text{mol}} \approx 1.27$ $\text{Ratio of H} = \frac{0.3968\, \text{mol}}{0.3968\, \text{mol}} = 1$ $\text{Ratio of O} = \frac{0.4206\, \text{mol}}{0.3968\, \text{mol}} \approx 1.06$
Since 1.27 is close to 1.25 (or $1 \frac{1}{4}$ ), we multiply all ratios by 4 to get whole numbers.
$\text{C ratio} = 1.27 \times 4 = 5.08 \approx 5$ $\text{H ratio} = 1 \times 4 = 4$ $\text{O ratio} = 1.06 \times 4 = 4.24 \approx 4$
Write the empirical formula.
The empirical formula is $\boxed{\text{C}_5\text{H}_4\text{O}_4}$ .

5) A compound contains C = 48g, H = 8g, and O = 12g.
Convert masses to moles.
$\text{Moles of C} = \frac{48\, \text{g}}{12.01\, \text{g/mol}} = 3.997\, \text{mol}$ $\text{Moles of H} = \frac{8\, \text{g}}{1.008\, \text{g/mol}} = 7.936\, \text{mol}$ $\text{Moles of O} = \frac{12\, \text{g}}{16.00\, \text{g/mol}} = 0.750\, \text{mol}$
Divide by the smallest number of moles to find the mole ratio.
$\text{Ratio of C} = \frac{3.997\, \text{mol}}{0.750\, \text{mol}} \approx 5.33$ $\text{Ratio of H} = \frac{7.936\, \text{mol}}{0.750\, \text{mol}} \approx 10.58$ $\text{Ratio of O} = \frac{0.750\, \text{mol}}{0.750\, \text{mol}} = 1$
Since 5.33 is close to $5 \frac{1}{3}$ and 10.58 is close to $10 \frac{2}{3}$ , we multiply all ratios by 3 to get whole numbers.
$\text{C ratio} = 5.33 \times 3 = 15.99 \approx 16$ $\text{H ratio} = 10.58 \times 3 = 31.74 \approx 32$ $\text{O ratio} = 1 \times 3 = 3$
Write the empirical formula.
The empirical formula is $\boxed{\text{C}_{16}\text{H}_{32}\text{O}_3}$ .

Calculate the empirical formula of organic compound containing C= 61.6% and H = 12.8%

Step-by-step solution

Convert masses to moles.

Divide by the smallest number of moles to find the mole ratio.

Write the empirical formula.

Assume a 100 g sample.

Convert masses to moles.

Divide by the smallest number of moles to find the mole ratio.

Write the empirical formula.

Calculate the percentage of H.

Convert masses to moles.

Divide by the smallest number of moles to find the mole ratio.

Write the empirical formula.

Convert masses to moles.

Divide by the smallest number of moles to find the mole ratio.

Write the empirical formula.

Convert masses to moles.

Divide by the smallest number of moles to find the mole ratio.

Write the empirical formula.

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