Write formulas for the following compounds.
1)
Nitrogen tri-iodide
NI3
2)
Palladium (IV) sulfide
PdS2
3)
Rhenium (VI) phosphate
Re(PO4)2
4)
Hydroiodic Acid
HI
5)
Gallium oxalate
Ga2(C2O4)3
Name the following compounds.
6)
N2S3
diNitrogen trisulfide
7)
PoN2
Polonium (VI) nitride
8)
HClO3
Chloric Acid
9)
Al2(Cr2O7)3
Aluminum dichromate
10)
V2Se3
Vanadium (III) selenide
11)
For the hydrate Ce(SO4)29 H2O determine:
A) The formula mass of hydrate. 494.377g/mol
Ce(SO4)2 = 140.115 + 2(32.065) + 8(15.994) = 332.240 g/mol
9 H2O = 18(1.0079) + 9(15.9994) = 162.1368 g/mol
Total Hydrate = 332.240 + 162.1368 = 494.377g/mol
B) the percentage of water present. 32.7962%
%H2O = (162.1368/494.377) x 100 = 32.7962%
C) the amount of water that would be driven off if a 67.98 g sample were heated until it
was dry. 22.29 g
Mass H2O = (% x whole)/100 = (32.7962 x 67.98g)/100 = 22.29 g
D)
the amount of dry salt that would remain. 45.68g
Amt Salt = 67.98 g – 22.29 g = 45.68 g
12)
A sample of an Arsenic iodine compound has a mass of 893.07 g. The compound is
made using 689.5 g of Iodine. Its molecular mass is 657.4600 g/mol
A) Amount of Arsenic in the compound 203.6g
As = 893.07g – 689.5g = 203.6g
B) percentage of Arsenic in the compound 22.79%
%As = (203.6/893.07) x 100 = 22.79%
C) percentage of Iodine in the compound 77.20% or 77.21%
%I = (689.5/893.07) x 100 = 77.20% or 100 – 22.79 = 77.21%
D) number of moles of Arsenic 2.718 mol
203.6g x 1 mol/74.9216 g = 2.718 mol As
E) number of moles of Iodine 5.433 mo
689.5 g x 1 mol/126.9045 g = 5.433 mol I
F) Empirical formula AsI2
𝐀𝐬: 𝐈
𝟐. 𝟏𝟕𝟏𝟖 𝟓. 𝟒𝟑𝟑
:
𝟐. 𝟏𝟕𝟏𝟖 𝟐. 𝟏𝟕𝟏𝟖
𝟏: 𝟐
G) Formula Mass from the Empirical Formula 328.7306 g/mol
74.9216 + 2(126.9045) = 328.7306 g/mol
H) Molecular formula As2I4
X = molecular mass/empirical mass = 689.5 g/328.7306 g = 2
I) Name
diArsenic tetra-iodide
2(AsI2) = As2I4