Percent hydrate
Name: ______Date: ______Period: ______
Hydrates Worksheet
1) How is a hydrate different from other chemical compounds?
It has water molecules loosely attached to it. These water molecules can typically be removed through heating (a process called “dehydration”. Hydrates usually involve ionic compounds with transition metals as the cation.
2) Define the following terms:
· anhydrate
A molecule which has no water molecules attached to it. This term is usually only used when describing chemicals which have specifically had their water molecules removed during heating – in these cases, the word “anhydrate” is added to the name.
· dehydration
The process of removing water from a hydrate, usually through applied heat.
3) Name the following compounds:
a) FeCl3. 6 H2O _____ iron (III) chloride hexahydrate ______
b) CuSO4 . 5 H2O __copper (II) sulfate pentahydrate ______
4) Write the formulas for the following compounds:
a) barium chloride dihydrate ______ BaCl2 . 2 H2O______
b) magnesium sulfate heptahydrate _____ MgSO4 . 7 H2O ______
5) What is the percent composition of water in the compound in problem 4b?
24.31+32.07+4(16.00) + 7(18.02) = 246.52
7(18.02)/126.14 = 126.14 / 246.52 = 0.5117
0.5117 x 100 = 51.17%
6) If 125 grams of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain?
51.17% of 125 g = water
0.5117 x 125 g = 63.96 g of water
125 g – 63.96 = 61.04g
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Composition of hydrates
A hydrate is an ionic compound with water molecules loosely bonded to its crystal structure. The water is in a specific ratio to each formula unit of the salt. For example, the formula CuSO4 × 5H2O indicates that there are five water molecules for every one formula unit of CuSO4. Answer the questions below.
1. What percentage of water is found in CuSO4×5H2O?
63.55 + 32.07 + 4(16.00) + 5(18.02) = 249.7 amu
5(18.02) / 249.7 = 90.1 / 249.7 = 0.361
0.361 x 100 = 36.1%
2. What percentage of water is found in Na2S ×9H2O?
2(22.99) + 32.07 + 9(18.02) = 240.2
9(18.02) / 240.2 x 100 = 67.52%
3. A 5.0 g sample of a hydrate of BaCl2 was heated, and only 4.3 g of the anhydrous salt remained. What percentage of water was in the hydrate?
(5.0 g – 4.3 g)/5.0 g x 100 = 14%
4. A 2.5 g sample of hydrate of Ca(NO3)2 was heated, and only 1.7 g of the anhydrous salt remained. What percentage of water was in the hydrate?
(2.5g – 1.7g) / 2.5g x 100 = 32%
5. A 3.0 g sample of Na2CO3 × H2O is heated to constant mass. How much anhydrous salt remains?
2(22.99) + 12.01 + 3(16.00) + 18.02 = 124.01 g
18.02 / 124.01 x 100 = 14.53% water
0.1453 x 3.0 g = .44 g
3.0 – 0.44 = 2.6g
6. A 5.0 g sample of Cu(NO3)2×nH2O is heated, and 3.9 g of the anhydrous salt remains. What is the value of n?
(5.0g-3.9g)/5.0g x100 = 22% water Molar mass = 63.55 + 2(14.01) + 6(16.00) + n(18.02) = 187.57 + 18.02n
18.02n / (187.57 + 18.02n) = 0.22 41.3 + 3.96n = 18.02n
0.22(187.57 + 18.02n) = 18.02n 41.3 = 14.06n n=3
Or do it like empirical formula 3.9gCu(NO3)2 / 187.57g = 0.021 / 0.021 = 1
1.1g H2O /18.02 = 0.061 / 0.021 = 3
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7. From the following data calculate the percent hydrate:
Mass of crucible = 51.0g
Mass of crucible and sample before heating = 63.0 g – 51.0g = 12.0
Mass of crucible and sample after heating = 60.0 g - 51.0g = 9.0
(12.0 – 9.0) / 12.0 x 100 = 25% hydrate
8. From the following data calculate the percent hydrate:
Mass of crucible = 1.610 g
Mass of crucible and sample before heating = 1.73 g – 1.610 g =0.12 g
Mass of crucible and sample after heating = 1.70 g - 1.610 g = 0.09 g
(0.12 g – 0.09 g) / 0.12g x 100 = 25% hydrate
Review Problems
1. Calculate the percent composition of EACH ELEMENT in the compound Fe2O3?
2(55.85) + 3(16.00) = 159.7 amu
2(55.85)/159.7 x 100 = 69.94 % Fe 3(16.00)/159.7 x 100 = 30.06% O
2. What is the percent composition of each element in Sucrose (C12H22O11)?
12(12.01) + 22(1.01) + 11(16.00) = 342.34amu 12(12.01)/342.34 x 100 = 42.10% C
22(1.01)/342.34 x 100 = 6.49% H 11(16.00)/342.34 x 100 = 51.41% O
3. What is the percent composition of copper and oxygen in CuxOy, if you have the following lab data?
Mass of crucible = 40.0 g, Mass of crucible and sample before heating = 200.0 g, Mass of crucible and sample after heating = 167.0 g CuxOy ® Cu + O
Mass of CuxOy = 200.0 g –40.0g = 160.0 g % Cu = 127.0g / 160.0g x 100 = 79.38%
Mass of Cu = 167.0 g – 40.0 g = 127.0 g % O = 33.0/160.0 x 100 = 20.6 %
4. What is the empirical formula of a compound made of FexOy, if you have the following lab data?
Mass of crucible = 40.0g, Mass of crucible and sample before heating 360.0 g, Mass of crucible and sample after heating = 264.0 g
Mass of FexOy = 360.0g – 40.0g = 320.0 g
Mass of Fe = 264.0g - 40.0g = 224.0 g / 55.85 = 4.01 / 4.01 = 1 Fe2O3
Mass of O = 320.0 g – 224.0 g = 96.0 g / 16.00 = 6 / 4.01=1.5
5. Calculate the average atomic mass of copper, if 69.2% is copper 62.9 amu and 30.8% is copper 64.9 amu.
(0.692)(62.9amu) + (0.308)(64.9 amu) = 63.5 amu
6. How many atoms of CARBON are there in 500.0g of sucrose (C12H22O11)?
7. A compound is found to contain 43.67% of P and 56.33% of O, what is the empirical formula?
P = 43.67 / 30.97 = 1.41/1.41 = 1 P2O5
O = 56.33 / 16.00 = 3.52 / 1.41 = 2.5
8. A compound contains 112.3 g of Fe and 48.0 g of O, what is its empirical formula?
Fe = 112.3g/55.85 = 2.01/2 = 1
O = 48.0g/16.00=3/2=1.5 Fe2O3
9. How many amu are there in (NH4)3PO4?
3(14.01) + 12(1.01) + 30.97 + 4(16.00) = 149.12 amu
10. Calculate the average atomic mass of cobalt, if 30.0% is 59.10 amu, 41.0% is 58.93amu, and 29.0% is 58.81 amu.
(0.30)(59.10)+ (0.41)(58.93) + (0.29)(58.81) = 58.95 amu
11. Determine the molecular formula of a compound with an empirical formula of CH4 and a formula mass of 80.020 amu.
formula mass of CH4 = 12.01 + 4(1.01) = 16.05 amu C5H20
80.020 amu / 16.05 amu = 5
12. A sample of a compound with a formula mass of 994.5 amu is found to consist of 0.115 g Na and 0.1775g of Cl. Find its molecular formula.
Na = 0.115g/22.99 = 0.005 / 0.005 = 1 Empirical formula NaCl = 22.99 + 35.45 = 58.44amu
Cl = 0.1775g / 35.45 = 0.005/0.005 = 1 994.5 amu / 58.44amu=17 Na17Cl17
13. What is the empirical formula of a compound made of FexOy, if you have the following lab data? Mass of crucible = 50.0 g, mass of crucible and sample before heating = 370.0 g, mass of crucible and sample after heating 274.0g FexOy® Fe + O
FexOy = 370.0 g – 50.0 g = 320.0 g Fe2O3
Fe = 274.0 g – 50.0 g = 224.0 g / 55.85 = 4/4 = 1
O = 320.0 – 224.0 = 96.0 g / 16.00 = 6 / 4 = 1.5
Review % composition, empirical formula, molecular formula, and moles
1. What is the molar mass of (NH4)3CrO4? (170g)
3(14.0) + 12(1.01) + 52.00 + 4(16.00) = 170.15 g
2. How many grams are in 1 mol of Al2O3? (102g)
2(26.98) + 3(16.00) = 101.96 g
3. What is the % composition of sulfur in H2SO3? (39.0%)
2(1.01) + 32.07 + 3(16.00) = 82.09 g
32.07 / 82.09 x 100 = 39.06 %
4. Calculate the mass of 1 mole of H2SO4. (98.0g)
2(1.01) + 32.07 + 4(16.00) = 98.09 g
a. Calculate how many formula units.
b. Calculate how many ions.
5. How many moles is 1.204 x 1024 atoms?
6. What is the empirical formula for a compound made of C and H if there is 50% carbon by mass in every 100g of the compound? (CH12)
C = 50g / 12.01 = 4.16 / 41.6 = 1 CH12
H = 50g / 1.01=49.5/4.16 = 11.9 = 12
7. What is the mass of 0.2 moles of K2Cr2O7? (58.8g)
2(39.10) + 2(52.00) + 7(16.00) = 294.2 g/mol
8. A compound has 37.7% Na, 22.95% Si, and 39.34% O, what is its empirical formula? (Na2SiO3)
Na = 37.7 / 22.99 = 1.64/0.817 = 2 O = 39.34/16.00 = 2.46/0.817 = 3
Si = 22.95/28.09 = 0.817/0.817 = 1 Na2SiO3
9. If 1 mole of HCl neutralizes 1 mol of NaOH, how many grams of HCl are needed to neutralize 40 grams
of NaOH? (36.5 g)
NaOH = 22.99+16.00+1.01=40.00
40 g of NaOH is equal to 1 mol so you need 1 mole of HCl = (1.01+35.45) = 36.46g
10. What is the percent composition of each substance in a mixture that has 30g of peanuts, 45g of cashews, 10g of Almonds, 20 g of pistachios? Total = 30+45+10+20 = 105 g
Peanut = 30/105 x 100 = 28.6% Pistachios = 20/105 x 100 = 19.0%
Cashews = 45/105 x 100 = 42.9%
Almonds = 10/105 x 100 = 9.5%
11. A sample of a compound with a formula mass of 132.0 amu is found to consist of 0.28 g of Nitrogen and 0.16 grams of oxygen. Finds its molecular formula. (N6O3)
N = 0.28 g/14.00 = 0.02/0.01 = 2 N2O = (2(14.00) + 16.00) = 44 N6O3
O0.16/16.00 = 0.01 / 0.01 = 1 132.0 g/44g = 3
12. If you have 20 g of AlxOy, decompose into 10.59 g of Al, what is the empirical formula? (Al2O3)
AlxOy® Al + O Al: 10.59/26.98 = .393/.393 =1
20 g 10.59g 9.41g O: 9.41 / 16.00 = .588 / .393 = 1.49
Al2O3
13. What is the percent hydrate of BaCl2 ×2H2O?
137.33 + 2(35.45) + 2(18.02) = 244.27
36.04 / 244.27 x 100 = 14.75%
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