A 6.67 mg sample of a compound containing carbon, hydrogen, and oxygen underwent

Question

A 6.67 mg sample of a compound containing carbon, hydrogen, and oxygen underwent combustion producing 15.17 mg of CO2 and 6.19 mg of H2O. What is the empirical formula of the compound? (Type your answer using the format CxHyOz for the compound CxHyOz)
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(b) A compound is composed of elements C, H, and N. A 7.875-mg sample of the unknown compound was combusted, producing 21.363 mg of CO2 and 6.125 mg of H2O. What is the empirical formula for this compound? (Type your answer using the format CxHyNz for the compound CxHyNz)
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If the compound has a molar mass of 160

Explanation / Answer

carbon in 15.17mg CO2=(12/44)*15.17=4.137 mg =0.345*10^-3mole

hydrogen in 6.19mg H2O=(2/18)*6.19=0.688 mg =0.688*10^-3mole

oxygen=6.67 - 4.137 - 0.688 = 1.84 mg =0.115*10^-3mole

molar ratio of C : H : O = 0.345 : 0.688 : 0.115

smallest number 0.115
divide the ratio by the smallest number we get
molar ratio of C : H : O = 3:6:1

compound has empirical formula C3H6O

2) c in 21.363 mg of CO2=5.826mg=0.4855*10^-3mole

H in 6.125 mg of H2O =0.681mg=0.681*10^-3 mole

N=7.875-(5.826+0.681)=1.368mg=0.098*10^-3mole

C : H : N =0.486 : 0.681 : 0.098

or C : H : N =5 : 7 : 1

compound= C5H7N

x*(60+7+14)=160

x=2

molecular formula=C10H17N2

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