blanks provided 1. The empirical formula of an unknown substance is Ch E/mol wha
ID: 1042836 • Letter: B
Question
blanks provided 1. The empirical formula of an unknown substance is Ch E/mol what is the molecular formula of the substance. Molecular formula:_ A compound has the following percent composition: carbon, oxygen. a. Calculate the empirical formula for the compound 2. 19% hydrogen, 28.85% nitrogen, and rest is Empirical jormula b. Given that the molar mass of the substance is 194.19 g/mol, find its molecular formula. Molecular formula: 3. Balance the following equations: a HCI (aq)Mn02(s)- CaSO +HPO d. aluminum iron (I) sulfide iron metal+ aluminum sulfideExplanation / Answer
1)emperical formula = CHCl
emperical formula mass = 12.0+1.0+35.5=48.5
Molar mass= 290.9gram/mole
n=molar mass/emperical formula mass
n= 290.8/48.5=5.99=6.0
Molecular formula = nxemperical formula
molecular formula= 6xCHCl = C6H6Cl6.
2) % by mass of C=5.19%
%by mass of H=28.85%
% by mass of O= 100-[% by mass of C + % by mass of H]
= 100-34.04=65.96%
element % by mass atomic weight relative number simple ratio
C 5.19 12.0 5.19/12.0=0.4325 0.4325/0.4325=1.0
H 28.85 1.0 28.85/1.0=28.85 28.85/0.4325=66.7
O 65.96 16.0 65.96/16.0=4.1225 4.1225/0.4325=9.5
simple ratio
C 1.0x6= 6.0
H 66.7x6=400.2
O 9.5x6= 57
emperical formula = C6H400O57
emperical formula mass=1384
molar mass= 194.19 gram
3)
a) 4 HCl + MnO2 --------- MnCl2 + 2 H2O + Cl2
b) 2 As(OH)3 + 3 H2SO4 ------------ As2(SO4)3 + 6 H2O
c) Ca3(PO4)2 + 3 H2SO4 --------------- 3 CaSO4 + 2 H3PO4
d) 2 Al + 3FeS --------- 3 Fe + Al2S3
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