B. Percentage Composition1. Calculate the percent composition of sucrose ( C12H22011)%O = 176.00 g/mol x 100 = 51.41%342.34 g/molO- 11x 16.00 = 176.00 g/mol% H = 22.22 g/mol x 100 = 6.49%342.34 g/molH-22 x 1.0122.22 g/mol%3DC- 12 x 12.01 g 144.12 g/mol= 342.34 g/mol%C = 144.12 g/mol x 100 = 42.10%342.34 g/molTotal100 %Hint: Use the atomic mass of elements in the table below (taken from the PeriodicTable of Elements) in solving for Molar Mass and Percentage Composition.ElementAtomic MassElementAtomic MassAl26.98 amuCI35.45 amu12.01 amuFe55.84 amu14.01 amuN.Mg24.31 amuRWB 982L

Known: Atomic mass of O = 16.00 amu Atomic mass of H = 1.01 amu Atomic mass of C = 12.01 amu

1. Calculate the percent composition of sucrose ( C12H22011) %O = 176.00 g/mol x 100 = 51.41% 342.34 g/mol O- 11 x 16.00 = 176.00 g/mol 22.22 g/mol %H = 22.22 g/mol x 100 = 6.49% 342.34 g/mol H- 22 x 1.01 %3D C- 12 x 12.01 g 144.12 g/mol = 342.34 g/mol % C = 144.12 g/mol x 100 = 42.10% 342.34 g/mol %3D Total 100 % Hint: Use the atomic mass of elements in the table below (taken from the Periodic Table of Elements) in solving for Molar Mass and Percentage Composition. Element Atomic Mass Element Atomic Mass Al 26.98 amu CI 35.45 amu C. 12.01 amu Fe 55.84 amu 18:88 8MB 14.01 amu Mg 24.31 amu

1. Calculate the percent composition of sucrose ( C12H22011) %O = 176.00 g/mol x 100 = 51.41% 342.34 g/mol O- 11 x 16.00 = 176.00 g/mol 22.22 g/mol %H = 22.22 g/mol x 100 = 6.49% 342.34 g/mol H- 22 x 1.01 %3D C- 12 x 12.01 g 144.12 g/mol = 342.34 g/mol % C = 144.12 g/mol x 100 = 42.10% 342.34 g/mol %3D Total 100 % Hint: Use the atomic mass of elements in the table below (taken from the Periodic Table of Elements) in solving for Molar Mass and Percentage Composition. Element Atomic Mass Element Atomic Mass Al 26.98 amu CI 35.45 amu C. 12.01 amu Fe 55.84 amu 18:88 8MB 14.01 amu Mg 24.31 amu

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

Question

B. Percentage Composition
1. Calculate the percent composition of sucrose ( C12H22011)
%O = 176.00 g/mol x 100 = 51.41%
342.34 g/mol
O- 11x 16.00 = 176.00 g/mol
% H = 22.22 g/mol x 100 = 6.49%
342.34 g/mol
H-22 x 1.01
22.22 g/mol
%3D
C- 12 x 12.01 g 144.12 g/mol
= 342.34 g/mol
%C = 144.12 g/mol x 100 = 42.10%
342.34 g/mol
Total
100 %
Hint: Use the atomic mass of elements in the table below (taken from the Periodic
Table of Elements) in solving for Molar Mass and Percentage Composition.
Element
Atomic Mass
Element
Atomic Mass
Al
26.98 amu
CI
35.45 amu
12.01 amu
Fe
55.84 amu
14.01 amu
N.
Mg
24.31 amu
RWB 982L

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