A 2.230-g sample of a solid is subjected to combustion analysis, yielding 76.59%

Question

A 2.230-g sample of a solid is subjected to combustion analysis, yielding 76.59% C and 6 39% H. It may also contain oxygen. What is the empirical formula for this compound? The molar mass of ties compound is determined to be about 94 g/mol. What is the molecular formula for this compound? A 1.000-g sample of a liquid is subjected to combustion analysis, yielding 92.3% C and 7.7% H. It may or may not also contain oxygen. What is the empirical formula for this compound? The molar mass of this compound is determined to be about 78 g/mol. What is the molecular formula for this compound? A 2.000 g sample of a liquid compound that contains only carbon and hydrogen in its formula is subjected to combustion analysis. From the result, the lab determines that 0.2874 g of the original sample's mass is due to hydrogen. What is the percent by mass hydrogen for this sample? What is the empirical formula for this compound? The molar mass of this compound is determined to be about 71 g/mol. What is the molecular formula for this compound?

Explanation / Answer

The marked question is 91 so I am answering 91

91) mass of sample =1.00 g
%C = 92.3%

%H = 7.7%

% H + C = 100%
There is no Oxygen

92.3% of 1 g = 0.923 g = 0.923 / atomic mass = 0.923/ 12 = 0.07 mole Carbon

7.7% of 1 g = 0.077 g = 0.923 / atomic mass = 0.077/ 1 = 0.07 mole Carbon
molar ratio of C : H = 0.07 : 0.07
smallest number 0.07
divide the ratio by the smallest number we get
molar ratio of C : H = 1: 1

(a) empirical formula is CH

(b) molar mass = 78

CH = 12+1 = 13

number of CH unit = 78/13 = 6

so molecular formulae = C6H6

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