Name: Carmela Shane Villanueva
1. Calculate the molar mass of the following compounds: (2 pts each)
a. glauber’s salt (𝑵𝒂𝟐 𝑺𝑶𝟒 ∙ 𝟏𝟎𝑯𝟐 𝑶)
Molar Mass of 𝑁𝑎2 𝑆𝑂4
22.99 𝑔 𝑁𝑎
Na = (2 mol Na) ( 1 𝑚𝑜𝑙 𝑁𝑎 ) = 45.98 g Na
32.06 𝑔 𝑆
S = (1 mol S) ( 1 𝑚𝑜𝑙 𝑆 ) = 32.06 g S
Mass of 1 mol 𝑁𝑎2 𝑆𝑂4 + Mass
of 10 mol 𝐻2 𝑂
15.99 𝑔 𝑂
O = (4 mol O) ( 1 𝑚𝑜𝑙 𝑂 ) = 63.96 g O
Mass of 1 mol 𝑁𝑎2 𝑆𝑂4 = 142.00 g 𝑁𝑎2 𝑆𝑂4
Molar Mass of 10𝐻2 𝑂
142.00 g 𝑁𝑎2 𝑆𝑂4 + 180.10 g
𝐻2 𝑂 = 322.10 g 𝑵𝒂𝟐 𝑺𝑶𝟒 ∙
𝟏𝟎𝑯𝟐 𝑶
1.008 𝑔 𝐻
H = (2 mol H) ( 1 𝑚𝑜𝑙 𝐻 ) = 2.02 g H
15.99 𝑔 𝑂
O = (1 mol O) ( 1 𝑚𝑜𝑙 𝑂 )= 15.99 g O
Mass of 10 mol 𝐻2 𝑂 = 10(18.01) g 𝐻2 𝑂
= 180.10 g 𝐻2 𝑂
b. nickel(II) acetate tetrahydrate with the formula 𝐍𝐢(𝐎𝐂𝐎𝐂𝐇𝟑 )𝟐 ∙ 𝟒𝐇𝟐 𝐎
Molar mass of Ni(OCOCH3 )2
58.69 𝑔 𝑁𝑖
Ni = (1 mol Ni) ( 1 𝑚𝑜𝑙 𝑁𝑖 ) = 58.69 g Ni
15.99 𝑔 𝑂
O = (4 mol O) ( 1 𝑚𝑜𝑙 𝑂 ) = 63.96 g O
12.01 𝑔 𝐶
C = (4 mol C) ( 1 𝑚𝑜𝑙 𝐶 ) = 48.04 g C
1.008 𝑔 𝐻
H = (6 mol H) ( 1 𝑚𝑜𝑙 𝐻 ) = 6.05 g H
Mass of 1 mol Ni(OCOCH3 )2 +
Mass of 4 mol H2 O
176.74 g Ni(OCOCH3 )2 + 72.04
g H2 O
= 248.78 𝐍𝐢(𝐎𝐂𝐎𝐂𝐇𝟑 )𝟐 ∙
𝟒𝐇𝟐 𝐎
Mass of 1 mol Ni(OCOCH3 )2 = 176.74 g Ni(OCOCH3 )2
Molar Mass of 4H2 O
1.008 𝑔 𝐻
H = (2 mol H) ( 1 𝑚𝑜𝑙 𝐻 ) = 2.02 g H
15.99 𝑔 𝑂
O = (1 mol O) ( 1 𝑚𝑜𝑙 𝑂 )= 15.99 g O
Mass of 4 mol H2 O = 72.04 g H2 O
c. epsom salt hydrate (𝑴𝒈𝑺𝑶𝟒 ∙ 𝟕𝑯𝟐 𝑶)
Molar mass of 𝑀𝑔𝑆𝑂4
Mass of 1 mol 𝑀𝑔𝑆𝑂4 + Mass
of 7 mol of 𝐻2 𝑂
24.31 𝑔 𝑀𝑔
Mg = (1 mol Mg) ( 1 𝑚𝑜𝑙 𝑀𝑔 ) = 24.31 g Mg
32.06 𝑔 𝑆
S = (1 mol S) ( 1 𝑚𝑜𝑙 𝑆 ) = 32.06 g S
a. 120.33 g 𝑀𝑔𝑆𝑂4 + 126.07 g
𝐻2 𝑂 = 246.40 g 𝑴𝒈𝑺𝑶𝟒 ∙
𝟕𝑯𝟐 𝑶
15.99 𝑔 𝑂
O = (4 mol O) ( 1 𝑚𝑜𝑙 𝑂 ) = 63.96 g O
Mass of 1 mol 𝑀𝑔𝑆𝑂4 = 120.33 g 𝑀𝑔𝑆𝑂4
Molar mass of 𝐻2 𝑂
H = (2 mol H) (
1.008 𝑔 𝐻
1 𝑚𝑜𝑙 𝐻
) = 2.02 g H
15.99 𝑔 𝑂
O = (1 mol O) ( 1 𝑚𝑜𝑙 𝑂 )= 15.99 g O
Mass of 7 mol of 𝐻2 𝑂 = 126.07 g 𝐻2 𝑂
2. Problems on determination of number of molecules of water in a hydrate. (3
pts each)
a. A hydrate of Na2CO3 has a mass of 4.31 g before heating. After heating, the
mass of the anhydrous compound is found to be 3.22 g. Determine the formula of
the hydrate and then write out the name of the hydrate.
Since this is a hydrate, one must first determine the mass of water lost:
4.31 g - 3.22 g = 1.09 g 𝐻2 𝑂
Next, determine the molar mass of 𝑁𝑎2 𝐶𝑂3 and 𝐻2 𝑂
22.99 𝑔 𝑁𝑎
Na = (2 mol Na) ( 1 𝑚𝑜𝑙 𝑁𝑎 ) = 45.98 g Na
1.008 𝑔 𝐻
H = (2 mol H) ( 1 𝑚𝑜𝑙 𝐻 ) = 2.02 g H
12.01 𝑔 𝐶
C = (1 mol C) ( 1 𝑚𝑜𝑙 𝐶 ) = 12.01 g C
15.99 𝑔 𝑂
O = (1 mol O) ( 1 𝑚𝑜𝑙 𝑂 )= 15.99 g O
15.99 𝑔 𝑂
O = (3 mol O) ( 1 𝑚𝑜𝑙 𝑂 ) = 47.97 g O
Mass of 1 mol 𝐻2 𝑂 = 18.01 g 𝐻2 𝑂
Mass of 1 mol 𝑁𝑎2 𝐶𝑂3 = 105.96 g 𝑁𝑎2 𝐶𝑂3
Proceed to determining the moles of 𝑁𝑎2 𝐶𝑂3 and 𝐻2 𝑂 present
(3.22 g 𝑁𝑎2 𝐶𝑂3 ) (
1 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3
105.96 g 𝑁𝑎2 𝐶𝑂3
( 1.09 g 𝐻2 𝑂) (
1 𝑚𝑜𝑙 𝐻2 𝑂
18.01 g 𝐻2 𝑂
) = 0.03 𝑚𝑜𝑙 𝑁𝑎2𝐶𝑂3
) = 0.06 𝑚𝑜𝑙 𝐻2 𝑂
Find a whole number ratio:
0.03 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3
𝑁𝑎2 𝐶𝑂3 =
𝐻2 𝑂 =
0.03 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3
0.06 𝑚𝑜𝑙 𝐻2 𝑂
0.03 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3
The formula is:
=1
𝑵𝒂𝟐 𝑪𝑶𝟑 ∙ 𝟐𝑯𝟐 𝑶
Or sodium carbonate
dihydrate
=2
b. Given that the molar mass of Na2SO4·nH2O is 322.1 g/mol, calculate the
value of n. Write the formula and name of the hydrate.
Molar Mass of 𝑁𝑎2 𝑆𝑂4 is:
Na = (2 mol Na) (
22.99 𝑔 𝑁𝑎
1 𝑚𝑜𝑙 𝑁𝑎
32.06 𝑔 𝑆
S = (1 mol S) (
1 𝑚𝑜𝑙 𝑆
) = 45.98 g Na
) = 32.06 g S
The mass of 𝐻2 𝑂 is:
322.1 g - 142.0 g = 180.1 g
𝐻2 𝑂
O = (4 mol O) ( 1 𝑚𝑜𝑙 𝑂 ) = 63.96 g O
The moles of water is:
1 𝑚𝑜𝑙 𝐻 𝑂
(180.1 g 𝐻2 𝑂) (18.01 g 𝐻2 𝑂 ) =
Mass of 1 mol 𝑁𝑎2 𝑆𝑂4 = 142.0 g 𝑁𝑎2 𝑆𝑂4
10 mol 𝑯𝟐 𝑶
15.99 𝑔 𝑂
2
Thus: n = 10 and the formula is
𝑵𝒂𝟐 𝑺𝑶𝟒 ∙ 𝟏𝟎𝑯𝟐 𝑶
C. A 4.92-g sample of hydrated magnesium sulfate crystals (MgSO4·nH2O)
gave 2.40 g of anhydrous magnesium sulfate on heating to a constant mass.
Determine the value of n. Write the formula and name of the hydrate.
Since this is a hydrate, one must first determine the mass of water lost:
4.92 g − 2.40 g = 2.52 g 𝐻2 𝑂
Next, determine the molar mass of 𝑀𝑔𝑆𝑂4 and 𝐻2 𝑂
24.31 𝑔 𝑀𝑔
Mg = (1 mol Mg) ( 1 𝑚𝑜𝑙 𝑀𝑔 ) = 24.31 g Mg
32.06 𝑔 𝑆
S = (1 mol S) (
1 𝑚𝑜𝑙 𝑆
1.008 𝑔 𝐻
H = (2 mol H) ( 1 𝑚𝑜𝑙 𝐻 ) =
2.02 g H
15.99 𝑔 𝑂
O = (1 mol O) ( 1 𝑚𝑜𝑙 𝑂 )=
15.99 g O
) = 32.06 g S
15.99 𝑔 𝑂
O = (4 mol O) ( 1 𝑚𝑜𝑙 𝑂 ) = 63.96 g O
Mass of 1 mol 𝑀𝑔𝑆𝑂4 = 120.33 g 𝑀𝑔𝑆𝑂4
Mass of 1 mol 𝐻2 𝑂 = 18.01
g 𝐻2 𝑂
Proceed to determining the moles of 𝑁𝑎2 𝐶𝑂3 and 𝐻2 𝑂 present
(2.40 g 𝑀𝑔𝑆𝑂4 ) (
1 𝑚𝑜𝑙 𝑀𝑔𝑆𝑂4
120.33 g 𝑀𝑔𝑆𝑂4
( 2.52 g 𝐻2 𝑂) (
1 𝑚𝑜𝑙 𝐻2 𝑂
18.01 g 𝐻2 𝑂
) = 0.02 𝑚𝑜𝑙 𝑀𝑔𝑆𝑂4
) = 0.14 𝑚𝑜𝑙 𝐻2 𝑂
Find a whole number ratio:
𝑀𝑔𝑆𝑂4 =
𝐻2 𝑂 =
0.02 𝑚𝑜𝑙 𝑀𝑔𝑆𝑂4
0.02 𝑚𝑜𝑙 𝑀𝑔𝑆𝑂4
0.14 𝑚𝑜𝑙 𝐻2 𝑂
0.02 𝑚𝑜𝑙 𝑀𝑔𝑆𝑂4
=1
=7
The formula is:
𝑴𝒈𝑺𝑶𝟒 ∙ 𝟕𝑯𝟐 𝑶
Or magnesium sulfate
heptahydrate
Thus, n = 7