Please check my work.Analyze the data for the sample to get the Empirical formula of CxHyOzGiven data for sample compound with oxygen:Mass of sample (g): 11.98Mass of carbon dioxide collected (g): 24.21Mass of water collected (g): 8.68 MOI C = 24.210)o 1 MOI Co1 Mol eメ=.5501 Mol C* 44.0094)coa%3D1 Mol coagramb of C = 5501 MOIC12.011 (3)<6.607(3)c%3Dメ1 MOI CMoI H = 8.68(3) HaO 1 Mol HyO18.0148(1)402 MOI H= .964 MOl H1 MO H2Dgrams of H-.964 MOI H1.0079 (6)H= ,972 (9)H%3Dメ1 Md Hgrams of Oz 11 98 (4) - C6.6607(4)-.972(8)=4.401()0MOlo = 4.401(5)01 MOI O= ,2751 MOIOメ15.999(6)0C5501 964H.CHOa 3.5.2751%3D.27512751 .2751%3D2 3.5472CHO47 2

Answer: This question based on the principle that every compound always form in the fixed ratio of…

Please check my work. Analyze the data for the sample to get the Empirical formula of CxHyOz Given data for sample compound with oxygen: Mass of sample (g): 11.98 Mass of carbon dioxide collected (g): 24.21 Mass of water collected (g): 8.68

Please check my work. Analyze the data for the sample to get the Empirical formula of CxHyOz Given data for sample compound with oxygen: Mass of sample (g): 11.98 Mass of carbon dioxide collected (g): 24.21 Mass of water collected (g): 8.68

Organic Chemistry: A Guided Inquiry

2nd Edition

ISBN:9780618974122

Author:Andrei Straumanis

Publisher:Andrei Straumanis

ChapterL2: Mass Spectrometry

Section: Chapter Questions

Problem 20CTQ

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Concept explainers

Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can be used in many calculations related to mole calculation, molar amount, mole ratio, and so on, which are significant in chemistry.

Question

Please check my work.

Analyze the data for the sample to get the Empirical formula of CxHyOz

Given data for sample compound with oxygen:

Mass of sample (g): 11.98

Mass of carbon dioxide collected (g): 24.21

Mass of water collected (g): 8.68

MOI C = 24.210)o 1 MOI Co
1 Mol e
メ
=.5501 Mol C
* 44.0094)coa
%3D
1 Mol coa
gramb of C = 5501 MOIC
12.011 (3)<
6.607(3)c
%3D
メ
1 MOI C
MoI H = 8.68(3) HaO 1 Mol HyO
18.0148(1)40
2 MOI H
= .964 MOl H
1 MO H2D
grams of H-
.964 MOI H
1.0079 (6)H
= ,972 (9)H
%3D
メ
1 Md H
grams of Oz 11 98 (4) - C6.6607(4)-.972(8)=4.401()0
MOlo = 4.401(5)0
1 MOI O
= ,2751 MOIO
メ
15.999(6)0
C5501 964
H.
CHO
a 3.5
.2751
%3D
.2751
2751 .2751
%3D
2 3.5
472
CHO
47 2

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