Question

A compound contains only carbon, hydrogen, and oxygen. combustion of 11.75 mg of the compound yields 17.61 mg co2 and 4.81 mg h2o. the molar mass of the compound is 176.1 g/mol. what are the empirical and molecular formulas of the compound? (type your answer using the format co2 for co2.)

Answer 1

Number of moles is defined as the ratio of given  mass in g to the molar mass.

First, convert the given mass of carbon dioxide in mg to g:

1 mg = 0.001 g

17.61 mg = 0.01761 g

Number of moles of carbon dioxide = $\frac{0.01761 g}{44.01 g/mol}$

= $0.0004001 mol$

Mass of carbon  = number of moles of carbon dioxide \times molar mass of carbon

= $0.0004001 mol\times 12.011 g/mol$

= $0.004806 g$

Number of moles of water= $\frac{0.00481 g}{18 g/mol}$

= $2.672\times 10^{-4}$

Since, water contains two hydrogen atoms. Thus,

Moles of hydrogen = $2\times 2.672\times 10^{-4}$

= $5.34\times 10^{-4}$

Mass of hydrogen = $5.34\times 10^{-4}\times \times 1.008 g/mol$

= $5.34\times 10^{-4} g$

Mass of oxygen = $0.001175-(5.38\times 10^{-4}g+0.004806 g)$

= $0.006405 g$

Number of moles of oxygen = $\frac{0.006405 g}{15.999 g/mol}$

= $0.000400$

Now,

$C_{0.0004001} H_{0.000534} O_{0.000400}$

Divide the smallest number to get the whole number,

$C_{\frac{0.0004001}{0.000400}} H_{\frac{0.000534}{0.000400}} O_{\frac{0.000400}{0.000400}}$

we get,

$C_{1} H_{1.33} O_{1}$

Now, multiply all the subscript by 3 to get the whole number,

$C_{3} H_{4} O_{3}$   (empirical fomula)

Molar mass of the compound  =$3\times 12.011 g/mol+4\times 1.008 g/mol+3\times 15.999 g/mol$

= $88.062 g/mol$

Divide given molar mass of the compound with the molar mass of the compound.

=$\frac{176.1 g/mol}{88.062 g/mol}$

= $1.999\simeq 2$

Thus, multiply the subscripts of empirical formula by 2 to get the molecular formula, we get:

$C_{6}H_{8}O_{6}$

Hence, empirical formula is $C_{3}H_{4}O_{3}$ and molecular formula is $C_{6}H_{8}O_{6}$