A 100 mg sample of a compound containing C, H, and O is analyzed.All the carbon is converted to CO_2 with mass 176 mg.All hydrogen is converted to H_2O (water) with mass 36 mg.Goal: determine empirical formula, molecular formula if there are 8 carbon atoms per molecule, and molar mass.My work is attached but I am not understanding the last two parts.Thank you. 8.100.mg .sample... Cx. Hy OzCO₂w/ mass..mg.wl. mass36 mg.need empircal formula, molecular formula, molan massCO₂ = (2.01 + 2 (16.00) = 44,01All CAll Hmg36converted toconverted0.176 A.44.01.4..molXтуд х6..036. g.18.02..malH₂0 = 2(1.008) + 16.004.00 mot. C..2.00 myt HC: 48.04.2.016.10-3 g.=to1 mg.10-3H₂0..1.mg.0.00 3999 .=.4.00·0.00.19912.01 gmyt0.1761.0089mo!23.83 =2420.036=g.18.016emporal C 24 H₁ H₂4 C₁ →formala249mol.2.00.५४.०५1768. C.atams.g2.016x 10-³ mol.molecule.x 10-3 mal= 18.02 g (4 SF)gC. = mass.co.₂= 2.016 = 12,016molecular formulaH₂0. (.3 SF).of CH = mass of.H.(3 SF)molan mass.(6 (1.008) +12 (12.01) =160.2 amolWhy minus 8instead of dividing by 8?

We have to calculate empirical formula, molecular formula and molar mass.

Answered: A 100 mg sample of a compound…

Chemistry

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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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A 100 mg sample of a compound containing C, H, and O is analyzed. All the carbon is converted to CO_2 with mass 176 mg. All hydrogen is converted to H_2O (water) with mass 36 mg. Goal: determine empirical formula, molecular formula if there are 8 carbon atoms per molecule, and molar mass. My work is attached but I am not understanding the last two parts. Thank you.

8.
100
.mg .sample... Cx. Hy Oz
CO₂
w/ mass.
.mg
.wl. mass
36 mg.
need empircal formula, molecular formula, molan mass
CO₂ = (2.01 + 2 (16.00) = 44,01
All C
All H
mg
36
converted to
converted
0.176 A.
44.01.4.
.mol
X
туд х
6..036. g.
18.02.
.mal
H₂0 = 2(1.008) + 16.00
4.00 mot. C.
.
2.00 myt H
C: 48.04.
2.016
.
10-3 g
.
=
to
1 mg.
10-3
H₂0.
.1.mg
.
0.00 3999 .=.4.00
·0.00.199
12.01 g
myt
0.176
1.0089
mo!
23.83 =24
2
0.036
=
g.
18.016
emporal C 24 H₁ H₂4 C₁ →
formala
24
9
mol.
2.00.
५४.०५
176
8. C.atams.
g
2.016
x 10-³ mol.
molecule.
x 10-3 mal
= 18.02 g (4 SF)
g
C. = mass
.co.₂
= 2.016 = 1
2,016
molecular formula
H₂0. (.3 SF).
of C
H = mass of.H.
(3 SF)
molan mass.
(6 (1.008) +12 (12.01) =
160.2 a
mol
Why minus 8
instead of dividing by 8?

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