Chem 121 Problem set III Solutions - 1
Problem Set III Stoichiometry - Solutions
 55.85 g Fe 
2
1. mass Fe   2.25 mol Fe  
  1.26 x 10 g Fe
 1mol Fe 
 1mol Zn   6.022 x 1023 atoms Zn 
23
2. Zn atoms   20.0 g Zn  
  1.84 x 10 atoms Zn

1mol Zn
 65.38 g Zn 

3. molecular mass of ethane = 2(12.011) + 6(1.008) = 30.07 g
 1molecule ethane 
molecules ethane   50.3 g ethane  
  1.67 molecules
 30.07 g ethane 
4. molecular mass of aniline = 6(12.011) + 7(1.008) + 14.01= 93.14 g/mol
 6 x 12.01 g C 
mass C  125.00 g C6H7N 
  96.71 g C
 93.14 g C6H7N 
5.
Let the molar mass of X be x.
mass of O in 40.0mg X4 O10 =40.0mg 
1mol X4 O10
1g
10mol O 16.00g



 22.5 x 10-3mg
1000mg (4x+160)g 1mol X4 O10 1mol O
40.0  10  16.00
 22.5
(4x  160)
x = 31.1 g/mol and the element is P.
6.
Let atomic mass of M be x
mass of water formed from 0.642g hydrate =0.642g 
1mol MCl2 .2H2O
2mol H2O
18.016g


 0.0949g
(x+106.94)g
1mol MCl2 .2H2O 1mol H2O
0.642  1 2  18.016
 1.683
23.1325=0.0949x+10.1486
(x+106.94)  1 1
solving for x = 137 g/mol which is Ba
how would you check your answer?
7.
Let atomic mass of M be x
 2 M2O3
1mol M 2mol M2O3 (2x  48)g
moles of oxide formed from 1.443g metal =1.443g 


 1.683g
x
4mol M
1mol M2O3
4M
+
3 O2
1.443  1 2  (2x  48)
 1.683
5.772x+138.528=6.732x
x  4 1
solving for x = 144.3 g/mol which is Nd
how would you check your answer?
8.
Molecular mass of C2H5OH is (2x12.011 + 6x1.0079 + 15.999) = 46.069
%C 
2mol  12.011g mol
 100  52.144%
46.069 g mol
%H 
6mol  1.0079 g mol
 100  13.127%
46.069 g mol
%O=34.729%
Molecular mass of Ca(HCO3)2 is (40.078 + 2x1.0079 + 2x 12.011 + 6x15.999) = 162.11
%Ca = 24.72%,
%C = 14.82%,
%H = 1.24%,
%O = 59.22%
Chem 121 Problem set III Solutions - 2
Molecular mass of MgNH4PO4 is (24.305 + 14.007 + 4x1.0079 + 30.974 + 4x15.999) = 137.3
Mg = 17.70%, N = 10.20%,
H = 2.936%,
P = 22.56%,
O = 46.61%
9. - use CO2 to get mass of C and H2O to get mass of H
 1mol CO2  1mol C  12.01 g C 
mass C   2.23 g CO2  


  0.6086 g C
 44.01 g CO2  1mol CO2   1mol C 
 1mol H2O  2 mol H  1.008 g H 
mass H  1.37 g H2O  


  0.1533 g C
 18.02 g H2O  1mol H2O   1mol H 
 quick check to see that C and H are the only components of this compound:
0.6086 g + 0.1533 g = 0.7619 g
OK
 if this added up to < 0.761 g, we would know that there must a 3rd element present, for which we could get the
mass by difference




0.6086 g C
0.1533 g C
%C  
%H  
 100%   79.9 %C
 100%   20.1%H
 0.761 g compound 
 0.761 g compound 
10. The first step is to find the percentage composition of the unknown, as follows:
moles CO2  moles C in CO2  moles C in compound  mass C in compound  percent C
1 mole CO2
1 mole C
12.01 g C
mass of C in CxHy Oz = 0.1486 g CO2 x
x
x
= 0.040555 g C
44.098 g CO2
1 mole CO2
1 mole C
percent C =
0.040555 g C
x 100 = 40.0 % C
0.1014 g unknown
moles H2O  moles H in H2O  moles H in compound  mass H in compound  percent H
1 mole H2O
2 moles H
1.008 g H
mass of H in CxHy Oz = 0.0609 g H2 O x
x
x
= 0.00681 g H
18.016 g H2O 1 mole H2O
1 mole H
0.00681 g H
x 100 = 6.72 % H
0.1014 g unknown
Percentage of O in unknown = 100 - 40.0 - 6.72 = 53.3 %
percent H =
The next step is to find the empirical formula from the percentage composition. I will assume I have 100.0 g of
the unknown for convenience.
Element
Relative mass
C
40.0
H
6.72
O
53.3
Relative number of moles (atoms)
40.0
= 3.33
12.01
6.72
= 6.67
1.008
53.3
= 3.33
16.00
The empirical formula is CxHyOz = C1H2O1
or
Divide by the smallest number
3.33
= 1.0
3.33
6.67
= 2.0
3.33
3.33
= 1.0
3.33
CH2O
To find the molecular formula, the molecular mass needs to be known. Let's say that the molecular mass has
been measured by mass spectrometry and found to be 180.0 amu.
Since the empirical formula is CH2O, the empirical mass is 12.01 + 2 x 1.008 + 16.00 = 30.02 amu.
Molecular mass
180.0
The ratio of molecular mass to empirical mass is
=
= 6.0
Empirical mass
30.02
So the molecular formula is (CH2O)6 or C6H12O6.
Chem 121 Problem set III Solutions - 3
11.
Take a 100.00 g sample
Atom
Na
Al
F
Mass (g)
32.79
13.02
54.19
Moles
32.79 / 22.99 = 1.4263
13.02 / 26.982 = 0.48255
54.19 / 18.998 = 2.8523
Divide by smallest
2.956
1
5.911
Best ratio
3
1
6
Na3AlF6
12.
mass of O in Lax Oy =0.05254g H2O 
1 mole H2O
1 moles O
16.00g O

x
=0.046661g
18.016 g H2O 1 mole H2O
1 mole O
mass of La in Lax Oy  0.3167  0.046661  0.27004
Element
Relative mass
Relative number of
moles (atoms)
Divide by the smallest
number
La
0.27004
0.27004
= 1.9441 10-3
138.9
1.9441 10-3
= 1.0
1.9441 10-3
O
0.046661
0.046661
2.9163  10-3
= 2.9163  10-3
 1.500
16.00
1.9441 10-3
The empirical formula is LaxOy = La2O3
13. you don’t have to come up with an equation but it can help
Ti + S  TixSy
a) mass of product = 31.700 g – 11.120 g = 20.58 g product
100% of Ti used resulted in product formation
 the % of Ti in the product is:
8.820 g Ti
x 100%  42.86 %Ti
20.58 g product
and all the rest is S  100 – 42.86 = 57.14 %S
b. for the empirical formula, require moles
 1mol Ti 
  0.184 mol Ti
 47.90 g Ti 
 8.820 g Ti 
 sulphur was in excess,  use the % S in the product to get the amount of sulphur that must have reacted:
 57.14 g S 
 1mol S 

  20.58 g product  
  0.3668 mol S
 100 g product 
 32.06 g S 
or could have got g S from 20.58 g – 8.820 g = 11.76 g S
 1mol S 
and 11.76 g S  
  0.3668 mol S
 32.06 g S 
Ti0.184/0.184 S0.367/0.184  Ti1S1.99 ~ TiS2
Chem 121 Problem set III Solutions - 4
14.
Mo2O3  MoxOy
12.64g
13.48g
The increase in mass is oxygen, so mass of extra O is 13.48 - 12.64 = 0.84 g
1mol Mo2O3
3mol O
16.00g
but mass of O in Mo2O3  12.64g 


 2.529g
239.88g
1mol Mo2O3 1mol O
so total mass of oxygen in new oxide is 2.529g + 0.84g = 3.369g
so mass of Mo in MoxOy = 13.48g - 3.369g = 10.1107g
Element
Mass
Mo
O
10.111
3.369
Relative number of
moles (atoms)
0.105386
0.21056
Divide by the smallest
number
1
1.998
Whole number
ratio
1
2
Empirical formula of the oxide is MoO2
15.
N2O5 + H2O  2 HNO3
Mg2C3 + 4 H2O
 2 Mg(OH)2 + C3H4
PCl5 + 4 H2O  H3PO4 + 5 HCl
16 Cr(s) + 3 S8(s)  8 Cr2S3(s)
Au2S3(s) + 3 H2(g)  2 Au(s) + 3 H2S(g)
6 NH4ClO4(s) + 10 Al(s)  3 N2(g) + 6 HCl(g) + 9 H2O(g)
16. - first write what you know from the question (you should know that bromine molecules, in fact all the
halogens, exist as diatomic molecules)
Na + Br2  NaBr
- then balance it
2Na + Br2  2NaBr
17. - we know that when you combust a hydrocarbon, we get CO2 and H2O  and the oxidant in combustion is
always oxygen (almost always):
C4H10 + O2  CO2 + H2O
- now balance it:
C4H10 + 13/2O2  4CO2 + 5H2O
- but we don’t want fractional coefficients:
2C4H10 + 13O2  8CO2 + 10H2O
18. - 1st we write what we know in an equation:
NH3 + CuO  N2 + Cu
- 2nd we balance it  but wait, we’ve only got H and O on the reacting side  there must be the formation of
water in this reaction  can balance it now:
2NH3 + 3CuO  N2 + 3Cu + 3H2O
- can now do a  c
 3 mol CuO 
a) mol CuO   0.445 mol NH3  
  0.668 mol CuO
 2 mol NH3 
 1mol N2   28.02 g N2 
b) mass N2   3.18 mol CuO  
  29.7 g N2

 3 mol CuO  1mol N2 
 1mol Cu  2 mol NH3   17.04 g NH3
b) mass NH3   55.0 g Cu  


 63.55 g Cu  3 mol Cu   1mol NH3

  9.83 g NH3

Chem 121 Problem set III Solutions - 5
19. mass of W formed = 100.0 g WO3 x
mass of H2O formed = 100.0 g WO3 x
1 mole WO3
1 mole W
183.9 g W
x
x
= 79.3 g W
231.9 g WO3
1 mole WO3
1 mole W
1 mole WO3
3 mole H2O
18.02 g H2O
x
x
= 23.3 g H2 O
231.9 g WO3
1 mole WO3
1 mole H2O
1 mole WO3
3 mole H2
2.016 g H2
x
x
= 2.61 g H2
231.9 g WO3
1 mole WO3
1 mole H2
CHECK: The sum of the masses of reactants = the sum of the masses of products
Reactants: 79.3 g + 23.3 g = 102.6 g
Products: 100.0 g + 2.61 g = 102.6 g
mass of H2 used = 100.0 g WO3 x
20.
6 NH4ClO4
+

10 Al
5 Al2O3
mass Al=5.0  103kg NH4 ClO4 
21.
2 C57H110O6
+
mass water=1.0kg fat 
22.

163 O2
+
3 N2
+
6 HCl
+
9 H2O
1000g 1mol NH4 ClO4
10mol Al
26.98g Al



 1.91 106 g
1kg
117.49g
6mol NH4 ClO4
1mol Al
114 CO2
+
110 H2O
1000g 1mol fat 110mol H2O 18.02g



 1 111g  1.1kg
1kg
891.5g
2mol fat
1mol H2O
Let mass BaO2 = (x) g , thus mass BaCO3 = (14.53 - x)g
Mass BaO formed:
1mol BaO2 1mol BaO
153.3g
from BaO2  x g BaO2 


 0.9055x g
169.3g
1mol BaO2 1mol BaO
from BaCO3  (14.53  x)g BaCO3 
1mol BaCO3
1mol BaO
153.3g


 0.77695(14.53  x) g
197.31g
1mol BaCO3 1mol BaO
but 12.37g BaO formed=[0.9055x+0.77695(14.53-x)]g
0.9055x  11.289  0.77695x  12.37
0.1285x  1.08097
x  8.409g
Percent BaO2 
23.
8.409g
 100  57.87%  57.9%
14.53g
2 NH3
+
3 CuO

N2
Mass nitrogen gas formed: from NH3 =18.1g NH3 
from CuO=90.4g CuO 
+
Percent BaCO3  42.1%
3 Cu
+
3 H2O
1mol NH3
1mol N2
28.013g


 14.88g
17.030g 2mol NH3 1mol N2
1mol CuO 1mol N2
28.013g


 10.61g
79.545g 3mol CuO 1mol N2
CuO is limiting, NH3 is in excess and 10.6g of nitrogen is formed
1mol CuO 2mol NH3 17.030g
Mass NH3 used=90.4g CuO 


 12.90g
79.545g 3mol CuO 1mol NH3
Mass NH3 remaining = 18.1 g - 12.9 g = 5.2 g
Chem 121 Problem set III Solutions - 6
24.
2 XeF2 + 2 H2O  2 Xe + 4 HF + O2
1.00g
50.0g
mass of HF formed
1mol XeF2
4mol HF
20.01g
from XeF2  1.00g 


 0.23639g
169.3g
2mol XeF2 1mol HF
1mol H2O 4mol HF
20.01g


 11 1.0g
18.02g
2mol H2O 1mol HF
thus XeF2 is limiting and H2O is in excess
mass HF formed is 0.236g
1mol XeF2 2mol H2O
18.02g
mass of H2O used=1.00g 


 0.1064g
169.3g
2mol XeF2 1mol H2O
from H2O  50.0g 
mass of H2O remaining
25.
=
4 NH3 (g)
2.00g
50.0g
-
+ 5 O2(g)
4.50g
0.1064g

=
49.9g
4 NO(g) +
6 H2O(l)
mass of NO formed
from NH3  2.00g 
26.
1mol NH3 4mol NO
30.01g


 3.524g
17.03g
4mol NH3 1mol NO
This problem can be solved in a couiple of ways, but the easiest is if it is treated as a empirical formula.
Percentage of MgSO4 in hydrate = 100  51.2 = 48.8
So take 100g of hydrate
molecule Mass (g)
Moles
Divide by
Closest whole number ratio
smallest
MgSO4
48.8
0.40545
1
1
H2O
51.2
2.84128
7.00
7
and so there are 7 molecules of water for every molecule of MgSO4 in the hydrate.
limiting reagent is AgNO3 , KCl is in excess and 4.2g AgCl is formed.
1mol AgNO3
1mol KCl
74.55g
mass KCl used  5.0g 


 2. 194g
169.91g
1mol AgNO3 1mol KCl
mass KCl remaining = 5.0g - 2.194g = 2.8g
1mol O2 4mol NO 30.01g
from O2  4.50g 


 3.376g
32.00g 5mol O2 1mol NO
thus O2 is limiting and NH3 is in excess
mass NO formed is 3.38g
1mol O2 4mol NH3
17.03g


 1.9 15g
32.00g 5 mol O2 1mol NH3
= 2.00g - 1.92g = 0.08g
Mass NH3 used  4.50g 
mass of NH3 remaining
27.
Percentage Yield = 1.21g * 100 / 1.25 g = 96.8%
28. - 1st write the equation:
N2 + H2  NH3
- then balance it:
N2 + 3H2  2NH3
- then determine limiting reactant:
Chem 121 Problem set III Solutions - 7
 with this easy an equation, can see that for 4 mol N2 would need 3 x 4 = 12 mol H2   H2 is limiting
- then calculate the theoretical yield on the basis of 6.0 mol H2:
 2 mol NH3 
mol NH3   6.0 mol H2  
  4.0 mol NH3
 3 mol H2 
 actual yield 
 1.6 
 then calculate %yield  
 100%   
 (100%)  40%
 4.0 
 theoretical yield 
C6H6 + Br2  C6H5Br + HBr
30.0g
65.0g
1mol C6H6 1mol C6H5Br
157.01g
mass of C6H5Br formed from C6H6  30.0g 


 60.30g
78.11g
1mol C6H6 1mol C6H5Br
29. a)
1mol Br2 1mol C6H5Br
157.01g


 63.86g
159.81g
1mol Br2
1mol C6H5Br
thus C6H6 is limiting and Br2 is in excess
mass C6H5Br formed is 60.3g
1mol C6H6
1mol Br2
159.81g
Mass Br2 used  30.0g 


 61.38g
78.11g
1mol C6H6 1mol Br2
mass of Br2 remaining = 65.0g - 61.4g = 3.6g
56.7g
b) Percentage Yield=
 100  94.0%
60.3g
from Br2  65.0g 
30. - 1st we write out the equation:
H2O + KO2  KOH + O2
- 2nd we balance it:
2H2O + 4KO2  4KOH + 3O2
- then we determine which reactant is limiting:
i) assume H2O is limiting
 1mol H2O  4 mol KOH  56.11 g KOH 
mol KOH   25.0 g H2O  


  156 g KOH
 18.02 g H2O  2 mol H2O   1mol KOH 
ii) assume KO2 is limiting:
 1mol KO2  4 mol KOH  56.11 g KOH 
mol KOH   25.0 g KO2  


  19.7 g KOH
 71.10 g KO2  4 mol KO2   1mol KOH 
 so KO2 is limiting since it forms the least amount of product
 and the theoretical yield = 19.7 g KOH
- 3rd calculate the % yield:
 15.3 g KOH actual 
%yield  
 (100%)  77.7%
 19.7 g KOH theoretical 
31. - we’re told how much iron is formed in the 2nd reaction
 we need to calculate the amount of CO required to get this amount of Fe, using the molar relationships from
the 2nd equation
 then we can calculate how much carbon is required to generate that much CO from the first equation
 1mol Fe  3 mol CO 
mol CO   750.0 g Fe  

  20.14 mol CO
 55.85 g Fe  2 mol Fe 
Chem 121 Problem set III Solutions - 8
 2 mol C  12.01 g C 
mass C   20.14 mol CO  

  241.9 g C
 2 mol CO  1mol C 
32.
 12.0 g NaCl 
mass NaCl  125.0 g soln  
  15.0 g NaCl
 100 g soln 
33.
 100 g soln  1mL soln 
mL soln   75.0 g NaCl  

  451mL soln
 15.0 g NaCl  1.108 g soln 
 1mol KNO3 
mol KNO3   3.765 g KNO3  
  0.03724 mol KNO3
 101.10 g KNO3 
 0.03724 mol KNO3  1000 mL soln 
[KNO3 ]  

  0.1241M KNO3
 300.0 mL soln  1L soln

34.
 1mol NaNO3 
mol NaNO3   20.00 g NaNO3  
  0.2353 mol NaNO3
 85.00 g NaNO3 
 1mL soln  1L

vol soln  100.0 g soln  

  0.08749 L soln
 1.143 g soln  1000 mL 
35.
[NaNO3 ] 
36.
0.2353 mol NaNO3
 2.689 M NaNO3
0.08749 L soln
M1V1 = M2V2
M2 =
M1V1
25.00 mL x 0.3447 M
=
= 0.00862 M
V2
1000.0 mL
 2.00 mol K 2SO4 
mol K 2SO4   0.0145 L  
  0.0290 mol K 2SO4
1L


0.0290 mol K 2SO4
[K 2SO4 ] 
 0.0967 M K 2SO4
0.300 L soln
37.
38. - we know M1 and V1 and M2
 we solve the dilution eqn for V2 and subtract the initial volume of HCl soln from final volume to determine the
amount of water to add:
M V 12.0 M 5.00 mL 
V2  1 1 
 100 mL soln
M2
 0.600 M
water added = 100 mL soln required – 5.00 mL 12.0 M HCl = 95 mL water added
39. - 1st calculate the # of moles of AgNO3 present
 then find the # of moles of K2CrO4 to react with it
 finally, use molarity to find the volume of interest
 0.420 mol K 2CrO4   2 mol AgNO3   169.98 g AgNO3 
mass AgNO3   75.00 mL K 2CrO4 soln  
  10.71 g AgNO3


 1000 mL soln  1mol K 2CrO4  1mol AgNO3 
moles MnO4 
1L
x 22.4 mL MnO4  x
= 4.48 x 10-4 mole
L
1000 mL
2+
5
mole
Fe
= 4.48 x 10-4 mole MnO4  x
= 2.24 x 10-3 mole
1 mole MnO4 
40. moles MnO4  reacted = 0.0200
moles of Fe2+ reacted with MnO4 
Chem 121 Problem set III Solutions - 9
[Fe2+ ] =
2.24 x 10-3 mole
40.0 mL Fe2+
x
1000 mL
= 0.056 M
1L
41.  determine the # mol KMnO4 , then mol Na2C2O4 required for the reaction
 then use definition of percentage
 0.08395 mol KMnO4  5 mol Na2C2O4 
mass Na2C2O4  18.74 mL KMnO4 soln  


 1000 mL KMnO4 soln  2 mol KMnO4 
 134.00 g Na2C2O4 
x
  0.5270 g Na2C2O4
 1mol Na2C2O4 
 0.5270 g Na2C2O4 
%Na2C2O4  
 100%   13.95 %Na2C2O 4
 3.778 g sample 
42.
H2SO4
2 NaOH

+ 2 H2O
1L
mol 1mol H2SO4
moles H2 SO4 =32.74mL NaOH 
 2.15

 0.035 195
1000mL
L
2mol NaOH
0.035195mol H2SO4 1000mL
mol
[H2SO4 ] 

 3.5195
 3.52M
10.00mL
1L
L
43. - convert mass KHP to moles KHP  which converts to mole NaOH and  M NaOH soln
 1mol KHP  1mol NaOH 
3
mol NaOH   0.7284 g KHP  

  3.567 x 10 mol NaOH
204.2
g
KHP
1mol
KHP



[NaOH] 
+
Na2SO4
3.567 x 10 3 mol NaOH
 0.1032 M NaOH
0.03458 L NaOH soln
2Cu2O  4Cu + O2
2CuO  2Cu + O2
44. Step 1.
Balance the equations
Step 2.
CuO is
Define the variables for CuO and Cu2O. Let the amount of Cu2O be X g and so the amount of
(1.000 - X) g.
Step 3:
Mass of Cu formed from:
1 mole Cu2O
4 mole Cu
63.55 g Cu
Cu2O: X g Cu2O x
x
x
= 0.88819X
143.1 g Cu2O
2 mole Cu2O
1 mole Cu
1 mole CuO
2 mole Cu
63.55 g Cu
x
x
= 0.79887(1.000-X)
79.55 g CuO
2 mole CuO
1 mole Cu
So the total mass of Cu formed is 0.88819X g + 0.79887(1.000-X) g = 0.8390 g
CuO: (1.000-X) g CuO x
Step 4: Solve for X
X = 0.4493 g = mass of Cu2O
0.4493 g
x 100 = 53.6 %
Therefore percent Cu2O in mixture is
0.8390 g
45.
Let mass of Mg = x , so mass of Zn = (1.000 - x) g
Mg + 1/2 O2  MgO
Zn + 1/2 O2  ZnO
mass of MgO from Mg=x g 
1mol Mg 1mol MgO
40.31g


 1.6582x g
24.31g
1mol Mg 1mol MgO
mass of ZnO from Zn=(1.000-x) g 
1mol Zn 1mol ZnO
81.38g


 1.2447(1.000x) g
65.38g
1mol Zn 1mol ZnO
Chem 121 Problem set III Solutions - 10
but mass of oxide mixture = 1.409 g = 1.6582x+1.2447(1.000-x)
1.2447  1.2447x  1.6582x  1.409
0.41346x  0.1643
x  0.3973g  0.397g
Percentage of Zn in the mixture is 60.3% , and percent Mg is 39.7%
46. - start with an unbalanced equation to get an idea of what’s going on:
KCl + MgCl2 + H2SO4  HCl
- we can write a balanced equation for the titration:
HCl + NaOH  H2O + NaCl
 we know that mol NaOH = mol HCl = mol Cl
 0.1054 mol NaOH 
3
mol Cl   75.82 mL NaOH 
  7.991 x 10 mol NaOH  mol Cl
 1000 mL NaOH soln 
 35.45 g Cl 
mass Cl  7.991 x 103 mol Cl 
  0.2833 g Cl
 1mol Cl 


- we also know that:
g KCl + g MgCl2 = 0.502 g
 so let x = g KCl, then g MgCl2 must be (0.502 –x)
 then we can calculate the mol of Cl that each compound contributes
 1mol KCl  1mol Cl  35.45 g Cl 
mass Cl from KCl   x g KCl  


  0.4762x g Cl
 74.55 g KCl  1mol KCl  1mol Cl 
 1mol MgCl2  2 mol Cl  35.45 g Cl 
mass Cl from MgCl2   x g MgCl2  


  0.7447(0.502  x) g Cl
 95.21 g MgCl2  1mol MgCl2   1mol Cl 
- the 2 compounds added together donated all the Cl in the HCl formed:

0.4762x + 0.7447(0.502 – x) = 0.2833 g Cl
0.4762x + 0.3738 - 0.7447x =0.2833
0.2685x = .09050
and x = 0.337 = g KCl
and  g MgCl2 = 0.502 – 0.337 = 0.165 g MgCl2