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The molar mass of silicon is 28.09 g/mol 28.1g SI x 1mol
= 1.0
28.09g
The molar mass of calcium is 40.08 g/mol
10.0g Ca x 1mol
= 0.245mol Ca
40.08g
0.250mol x 6.022x10
23
atoms/mol = 1.50x10
23
atoms
1.00mol H
2
O = 18.0g
10.0g x 1mol
= 0.556mol
18.0g
0.556mol x 6.022x10
23
molecules
1mol
=3.35x10
23
molecules
The molar mass of HNO
3
is 63.01 g/mol
1mol HNO
3
= 63.01g
3.26mol x 63.01g
= 205.4g
1 mol
Molar mass = the mass (in grams) of 1 mol of a substance 1mol CO
2
The formula of sodium carbonate is Na
2
CO
3
1mol Na
2
CO
3
= 2 mol Na, 1 mol C, 3 mol O
1 mol Na
2
CO
3
= 2(23.0g) + 12.0g + 3(16.0g) = 106g
The molar mass of Na
2
CO
3
is 106g
There are 2 moles of Na in Na
2
CO
3
Mass% Na = 2x23.0g
x 100% = 43.4% Na
106g
The molar mass of Na
2
CO
3
is 106g
There are 3 moles of O in Na
2
CO
3
Mass% O = 3x16.0g
x 100% = 45.3%
106g
The molar mass of Na
2
CO
3
is 106g
There is 1 mole of C in Na
2
CO
3
Mass% C = 1x12.0g
x 100% = 11.3%
106g
Chromium(III) oxalate is Cr
2
(C
2
O
4
)
3
MM = 2(52.00) + 6(12.01) + 12(16.00) = 368.1 g/mol
Ammonium chloride is NH
4
Cl
MM = 12.01 + 4(1.008) + 35.45 = 53.49 g/mol
1mol NH
4
Cl contains 1mol N
Mass% = 14.01g/mol
x 100 = 26.2%
53.49 g/mol
MM of Mg = 24.31 g/mol
0.450g x 1mol
= 0.0185mol
24.31g
Not elemental nitrogen (N
2
)
In a compound, N is bonded to other atoms, so we only consider the mass of the nitroge
MM of N = 14.01 g/mol
0.173g x 1mol
= 0.0123mol
12.01g
MM = 4(12.01) + 5(1.008) + 2(14.01) + 16.00
= 97.10 g/mol
Moles of Carbon
5.657g C x 1mol C
= 0.4714mol C
12.01g C
Moles of Hydrogen
0.3165g H x 1mol H
= 0.3140mol H
1.008g H
Moles of Chlorine
5.566g Cl x 1mol Cl
= 0.1570mol Cl
35.45g Cl
Divide through by smallest molar amount
0.4714mol C
= 3.003mol C
0.1570
0.3140mol H
= 2.000mol H
0.1570
0.1570mol Cl
= 1.000mol Cl
1.570
The empirical formula, CH
2
, has molar mass
12 + 2(1) = 14 g/mol
≈
The molar mass of the compound = 82 g/mol
The ratio of molar masses is:
82
= 5.9
6
≈
14
The molecular formula is 6 times the empirical formula
(CH
2
)
6
= C
6
H
12
Step1: assume 100.0g sample to find amounts and convert to moles of each element in the sample
12.1g C x 1mol
= 1.01mol C
12.01g
16.2g O x 1mol
= 1.01mol O
16.00g
71.7g Cl x 1mol
= 2.02mol Cl
35.34g
The molar ratios are essentially integer values, so no further manipulation is needed
%mass = mass
solute
x 100%
mass
solution
mass
solution
= mass
solvent
+ mass
solute
%mass =
2.50g
x 100%
50.0g + 2.50g
%mass = 4.76%
MM = 30.97 g/mol
+ 5(35.34 g/mol)
MM = 208.22 g/mol
10.5g x
1mol
= 0.0504mol
208.22g
MM = 208.22 g/mol
10.5g x
1mol
= 0.0504mol
208.22g
150.0mL x
1L
= 0.1500L
1000mL
M
= 0.0504mol
= 0.336
M
0.1500L
Molarity = mol
V
15.6g KBr x 1mol KBr
= 0.131mol
119.0g
Molarity = 0.131mol
= 0.105mol/L
1.25L
Mass% = mass
solute
x 100
mass
solution
mass
solution
= mass
solute
+ mass
solvent
mass
solute
= mass
solution
+ mass
solvent
= 153.4g – 125.2g = 28.2g
Mass% = 28.2g
x 100 = 18.38%
153.4g
Molarity (
M
) = mol
L
10.0g NaCl x
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This note was uploaded on 02/28/2008 for the course CHEM 1021 taught by Professor Hoenigman, during the Fall '07 term at Colorado.
 Fall '07
 HOENIGMAN,
 Chemistry, Atom, Mole

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