cive Lp?ResourcesA compound is 53.31% C, 11.18% H, and 35.51% O by mass. What is its empirical formula? Insert subscripts as needed.empirical formula: CHOThe molecular mass of the compound is 90 amu. What is the molecular formula? Insert subscripts as needed.molecular formula: CHO

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Asked Oct 23, 2019
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A compound is 53.31% C, 11.18% H, and 35.51% O by mass. What is its empirical formula? Insert subscripts as needed.
empirical formula: CHO
The molecular mass of the compound is 90 amu. What is the molecular formula? Insert subscripts as needed.
molecular formula: CHO
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cive Lp? Resources A compound is 53.31% C, 11.18% H, and 35.51% O by mass. What is its empirical formula? Insert subscripts as needed. empirical formula: CHO The molecular mass of the compound is 90 amu. What is the molecular formula? Insert subscripts as needed. molecular formula: CHO

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check_circleExpert Solution
Step 1

Given,

Mass percentage of C = 53.31 %

Mass percentage of H = 11.18 %

Mass percentage of O = 35.51 %

This means that 100 g of a compound contains 53.31 g of C, 11.18 g of H and 35.51 g of O.

Moles of C, H and O can be calculated as:

53.31 g
Mass
Moles of C
4.44 mol
Molar mass
12 g/mol
11.18 g
Mass
Moles of H =
11.18 mol
Molar mass
1 g/mol
35.51 g
Mass
Moles of O
= 2.22 mol
Molar mass
16 g/mol
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53.31 g Mass Moles of C 4.44 mol Molar mass 12 g/mol 11.18 g Mass Moles of H = 11.18 mol Molar mass 1 g/mol 35.51 g Mass Moles of O = 2.22 mol Molar mass 16 g/mol

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Step 2

The empirical formula can be given by calculating the simplest whole molar ratio as follows:

4.44 mol
2
2.22 mol
С -
11.18 mol
= 5
2.22 mol
Н-
2.22 mol
1
2.22 mol
The empirical formula = C2H50
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4.44 mol 2 2.22 mol С - 11.18 mol = 5 2.22 mol Н- 2.22 mol 1 2.22 mol The empirical formula = C2H50

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Step 3

The molecular formula of a comp...

Molecular formula = (Empirical formula)n
Molecular mass
Where, n
Empirical mass
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Molecular formula = (Empirical formula)n Molecular mass Where, n Empirical mass

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