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ladessa [460]
3 years ago
5

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 h2

o. 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.)
Chemistry
1 answer:
Ymorist [56]3 years ago
8 0

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}



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