Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 3, Problem 116AP
Interpretation Introduction

Interpretation:

The empirical formula and the molecular formula of the compound are to be determined with given molar mass.

Concept introduction:

The simplest atomic ratio of the mass of the elements present in a compound is known as the empirical formula.

The actual atomic ratio of the mass of the elements present in a compound is the molecular formula.

The mass of hydrogen in H2O

is

mass of H = wtofH2atomicmassofHmolecularmassofH2O

The mass of oxygen in H2O

is

mass of O = wtofH2atomicmassofOmolecularmassofH2O

The value of n is calculated as

n=molar massempirical formula mass

The molecular formula of the compound is given as

Molecular formula =Empirical formula×n

Expert Solution & Answer
Check Mark

Answer to Problem 116AP

Solution: The empirical formula and the molecular formula of the compound are CH2O and C6H12O6, respectively.

Explanation of Solution

Given information: The compound contains 40% carbon.

The hydrogen to oxygen ratio in the compound is 2:1.

The given compound contains 40% carbon, which means that in a 100 g sample, the mass of carbon is 40 g.

Therefore, the weight of the H2O in the compound will be (10040)=60 g.

The amount of H present in 60 g H2O is calculated as

18.0 g of water contains 2.0 g of hydrogen. Therefore, 60.0 g of water will contain:

mass of H = wtofH2atomicmassofHmolecularmassofH2O

60.0g×2.0g/mol18.0g/mol=6.67 g of hydrogen.

The amount of oxygen present in 60 g H2O is calculated as

18.0 g of water contains 16.0 g of oxygen. Therefore, 60.0 g of water will contain:

mass of O = wtofH2atomicmassofOmolecularmassofH2O

60.0g×16.0g/mol18.0g/mol=53.33 g of oxygen.

The formula for computing the number of moles of an element is as follows:

Numberofmoles =Given massAtomic mass …… (1)

Substitute the values of the given mass and the molar mass of the hydrogen atom in equation (1)

MolesofH=6.67g1.00g/mol=6.67 mol

Substitute the values of the given mass and the molar mass of the carbon atom in equation (1)

MolesofC=40.0g12.01g/mol=3.33 mol

Substitute the values of the given mass and the molar mass of the oxygen atom in equation (1)

MolesofO=53.33g16.00g/mol=3.33 mol

The simplest molar ratio between C, H, and O is 1:2:1 (obtained by dividing the values of moles by the least number, i.e., 3.33). Therefore, the empirical formula is CH2O.

The empirical formula mass of the compound is calculated as

Empiricalformulamass=(1×atomicmass of  C)+(2×atomicmass of H)+(1×atomicmassof  O)=(1×12.0g/mol)+(2×1.0g/mol)+(1×16.0g/mol)=30.0 g/mol

The value of n is calculated as

n=molar massempirical formula mass=178g/mol30g/mol=5.96

The molecular formula of the compound is given as

Molecular formula =Empirical formula×n=CH2O×6=C6H12O6

Conclusion

The empirical formula of the compound is CH2O.

The molecular formula of the compound is C6H12O6.

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Chapter 3 Solutions

Chemistry

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