   # A compound called MMT was once used to boost the octane rating of gasoline. What is the empirical formula of MMT if it is 49.5% C, 3.2% H, 22.0% O, and 25.2% Mn? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 2, Problem 133GQ
Textbook Problem
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## A compound called MMT was once used to boost the octane rating of gasoline. What is the empirical formula of MMT if it is 49.5% C, 3.2% H, 22.0% O, and 25.2% Mn?

Interpretation Introduction

Interpretation: The empirical formula of the given compound MMT should be determined using given data.

Concept introduction:

• Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.
• Equation for number moles from mass and molar mass,

Numberofmoles=MassingramsMolarmass

• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.

### Explanation of Solution

Given:

Mass percent of carbon, hydrogen, oxygen and manganese in the given compound MMT are 49.5%,3.2%,22.0%and25.2% respectively.

Mass percent of an element means, 100g of compound contains that mass percent of an element in grams. Therefore, mass of 49.5g carbon 3.2g of hydrogen 22.0g of oxygen 25.2g of manganese are present respectively in 100g of MMT.

Equation for number moles from mass and molar mass is,

Numberofmoles=MassingramsMolarmass

Therefore, the number of moles of carbon is,

Numberofmoles=49.5g12.01g/mol=4.1215mol

The number of moles of hydrogen is,

Numberofmoles=3.2g1g/mol=3.2mol

The number of moles of oxygen is,

Numberofmoles=22.0g16g/mol=1

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