Pro DCniel C. Harris - Q.. x100%SignatuEXAMPLECombustion Analysis CalculationsA compound weighing 5.714 mg produced 14.414 mg of CO, and 2.529 mg of H,O uponcombustion. Find the wt% of C and H in the sample.Solution One mole of CO, contains 1 mol of carbon. Therefore,Moles of C in samplemoles of CO, producedgto,14.414 X 10-343.275 X 10mol44.009 g/mol CO,Mass of C in sample(3.275 X 10 mol C)(12.010 6 g/mol C)= 3.934 mg3.934 mg Cwt% C =x 100 = 68.84%5.714 mg sampleOne mole of H,O contains 2 mol of H. Therefore.Moles of H in sample 2(moles of H,0 produced)g HO18.015 g/mol HO2.529 x 10- 2.808 × 10molMass of H in sample = (2.808 × 10 "mol H1.008 g/mol H) = 2.830 x 100.283 0 mg Hwt% H =x 100-495%5.714 mg sampleTEST YOURSELF A 6.234-mg sample produced 12.123 mg Co, and 2.529 mg of H.OFind the wt of C and H in the sample ARwer: 5307.4.54

Given mass of sample = 6.234 mg Mass of CO2 = 12.123 mg Mass of H2O = 2.529 mg

Answered: TEST YOURSELF A 6.234-mg sample…

TEST YOURSELF A 6.234-mg sample produced 12.123 mg CO. and 2.529 mg of H-O. Find the wt of C and H in the sample AnSwer: 53.07.454

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter31: Introduction To Analytical Separations

Section: Chapter Questions

Problem 31.19QAP

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Precipitation Titration

Precipitation analysis is used to analyze drugs, and to measure the salt content in food and beverages. It is an important concept from the examination point of view many practical questions are asked on the topic. This topic is significant in the professional exams for both undergraduate and graduate courses, especially for

Question

Pro DC
niel C. Harris - Q.. x
100%
Signatu
EXAMPLE
Combustion Analysis Calculations
A compound weighing 5.714 mg produced 14.414 mg of CO, and 2.529 mg of H,O upon
combustion. Find the wt% of C and H in the sample.
Solution One mole of CO, contains 1 mol of carbon. Therefore,
Moles of C in sample
moles of CO, produced
gto,
14.414 X 10-3
4
3.275 X 10
mol
44.009 g/mol CO,
Mass of C in sample
(3.275 X 10 mol C)(12.010 6 g/mol C)
= 3.934 mg
3.934 mg C
wt% C =
x 100 = 68.84%
5.714 mg sample
One mole of H,O contains 2 mol of H. Therefore.
Moles of H in sample 2(moles of H,0 produced)
g HO
18.015 g/mol HO
2.529 x 10
- 2.808 × 10mol
Mass of H in sample = (2.808 × 10 "mol H1.008 g/mol H) = 2.830 x 10
0.283 0 mg H
wt% H =
x 100-495%
5.714 mg sample
TEST YOURSELF A 6.234-mg sample produced 12.123 mg Co, and 2.529 mg of H.O
Find the wt of C and H in the sample ARwer: 5307.4.54

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