1  

Honors Chemistry Name _______________________ HW – Hydrates, Percent Composition, EF/MF Date ________________________ Complete these problems on a separate sheet of paper.

1. Determine the percent Oxygen in Potassium Perchlorate. KClO4 Molar Mass is 138.547 g KClO4 4 moles of O in KClO4 Molar mass of O is 15.999 4 moles O = 63.996 g

𝑃𝑒𝑟𝑐𝑒𝑛𝑡  𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛  𝑜𝑓  𝑂 =  63.996  𝑔  138.547  𝑔

 𝑥  100 =  46.19  %

2. Determine the percent Aluminum in Aluminum Sulfate.

Al2(SO4)3 Molar Mass is 342.15 g 2 moles of Al in Al2(SO4)3 Molar Mass of Al = 26.982 g 2 moles Al = 53.964 g

𝑃𝑒𝑟𝑐𝑒𝑛𝑡  𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛  𝑜𝑓  𝐴𝑙 =  53.964  𝑔342.15  𝑔

 𝑥  100 =  15.77  %

3. Determine the percent Iron in Iron II Phosphite.

Fe3(PO3)2 Molar Mass is 325.741 g 3 moles of Fe in Fe3(PO3)2 Molar Mass of Fe = 55.933 g 3 moles of Fe = 167.799 g

𝑃𝑒𝑟𝑐𝑒𝑛𝑡  𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛  𝑜𝑓  𝐹𝑒 =  167.799  𝑔325.741  𝑔

 𝑥  100 = 51.51%

4. A 4.829 gram sample of Magnesium Sulfate Septahydrate (Epsom Salt). If the sample is heated

and all of the water is driven off, how much anhydrous salt will remain?

𝐹𝑜𝑟𝑚𝑢𝑙𝑎  𝑜𝑓  𝐻𝑦𝑑𝑟𝑎𝑡𝑒 =  𝑀𝑔𝑆𝑂!   ⋅ 7𝐻!𝑂 𝐴𝑛ℎ𝑦𝑑𝑟𝑜𝑢𝑠  𝑆𝑎𝑙𝑡 =  𝑀𝑔𝑆𝑂!

4.829  𝑔  𝑀𝑔𝑆𝑂!   ⋅ 7𝐻!𝑂

1  𝑥  

1  𝑚𝑜𝑙𝑒  𝑀𝑔𝑆𝑂!   ⋅ 7𝐻!𝑂246.472  𝑔  𝑀𝑔𝑆𝑂!   ⋅ 7𝐻!𝑂

 𝑥  1  𝑚𝑜𝑙𝑒  𝑀𝑔𝑆𝑂!

1  𝑚𝑜𝑙𝑒  𝑀𝑔𝑆𝑂!   ⋅ 7𝐻!𝑂  𝑥  120.367  𝑔  𝑀𝑔𝑆𝑂!1  𝑚𝑜𝑙𝑒  𝑀𝑔𝑆𝑂!

=  2.358  𝑔  𝑀𝑔𝑆𝑂!

5. A 7.028 gram sample of a sodium sulfate hydrate is heated. After driving off all the water, 3.100 grams of the anhydrous salt is left. What is the formula of the hydrate?

𝐹𝑜𝑟𝑚𝑢𝑙𝑎  𝑜𝑓  𝐻𝑦𝑑𝑟𝑎𝑡𝑒 =  𝑁𝑎!𝑆𝑂!   ⋅ 𝑥𝐻!𝑂

𝐴𝑛ℎ𝑦𝑑𝑟𝑜𝑢𝑠  𝑆𝑎𝑙𝑡 =  𝑁𝑎!𝑆𝑂!

𝑀𝑜𝑙𝑒𝑠  𝑜𝑓  𝐴𝑛ℎ𝑦𝑑𝑟𝑜𝑢𝑠  𝑆𝑎𝑙𝑡 =3.100  𝑔  𝑁𝑎!𝑆𝑂!

1  𝑥  

1  𝑚𝑜𝑙𝑒  𝑁𝑎!𝑆𝑂!142.042  𝑔  𝑁𝑎!𝑆𝑂!

= 0.02182  𝑚𝑜𝑙𝑒𝑠  𝑁𝑎!𝑆𝑂!  

  2  

𝐺𝑟𝑎𝑚𝑠  𝑜𝑓  𝑊𝑎𝑡𝑒𝑟 = 7.028  𝑔 − 3.100  𝑔 = 3.928  𝑔  𝑊𝑎𝑡𝑒𝑟

𝑀𝑜𝑙𝑒𝑠  𝑜𝑓  𝑊𝑎𝑡𝑒𝑟 =3.928  𝑔  𝐻!𝑂

1  𝑥  

1  𝑚𝑜𝑙𝑒  18.015  𝑔  𝐻!𝑂

= 0.218  𝑚𝑜𝑙𝑒𝑠  𝐻!𝑂

𝑀𝑜𝑙𝑒𝑠  𝑜𝑓  𝐴𝑛ℎ𝑦𝑑𝑟𝑜𝑢𝑠  𝑆𝑎𝑙𝑡 ∶ 𝑀𝑜𝑙𝑒𝑠  𝑜𝑓  𝑊𝑎𝑡𝑒𝑟

0.02182  𝑚𝑜𝑙𝑒𝑠 ∶ 0.218  𝑚𝑜𝑙𝑒𝑠

𝐷𝑖𝑣𝑖𝑑𝑒  𝑏𝑦  𝑡ℎ𝑒  𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡  𝑛𝑢𝑚𝑏𝑒𝑟  𝑜𝑓  𝑚𝑜𝑙𝑒𝑠

0.021820.02182

∶  0.2180.02182

1 ∶ 10

𝐹𝑜𝑟𝑚𝑢𝑙𝑎  𝑜𝑓  𝐻𝑦𝑑𝑟𝑎𝑡𝑒 =  𝑁𝑎!𝑆𝑂!   ⋅ 10𝐻!𝑂

6. What is the empirical formula of a compound that contains 40.0% Carbon 6.7% hydrogen, and 53.3% Oxygen.

40.0  𝑔  𝐶

1  𝑥  

1  𝑚𝑜𝑙𝑒  𝐶12.011  𝑔  𝐶

=  3.3302  𝑚𝑜𝑙𝑒𝑠  𝐶

6.7  𝑔  𝐻

1  𝑥  

1  𝑚𝑜𝑙𝑒  𝐻1.008  𝑔  𝐻

=  6.6468  𝑚𝑜𝑙𝑒𝑠  𝐻

53.3  𝑔  𝑂

1  𝑥  

1  𝑚𝑜𝑙𝑒  𝑂15.999  𝑔  𝑂

=  3.3314  𝑚𝑜𝑙𝑒𝑠  𝑂

3.3302  𝑚𝑜𝑙𝑒𝑠  𝐶 ∶ 6.6468  𝑚𝑜𝑙𝑒𝑠  𝐻 ∶ 3.314  𝑚𝑜𝑙𝑒𝑠  𝑂

1: 2: 1 𝐶𝐻!𝑂

7. If the compound in number 6 has a molar mass of 180 grams/mole, what is the molecular

formula?

𝑀𝑜𝑙𝑎𝑟  𝑀𝑎𝑠𝑠  𝐸𝐹 = 30.026  𝑔

𝑋 =180  𝑔/𝑚𝑜𝑙𝑒

30.026  𝑔/𝑚𝑜𝑙𝑒= 6

𝐶!𝐻!"𝑂!

8. What is the empirical and molecular formula of a compound that contains 84% Carbon and 16%

Hydrogen and a molar mass of 114 g/mole.

84  𝑔  𝐶1

 𝑥  1  𝑚𝑜𝑙𝑒  𝐶12.011  𝑔  𝐶

=  6.9935  𝑚𝑜𝑙𝑒𝑠  𝐶

16  𝑔  𝐻1

 𝑥  1  𝑚𝑜𝑙𝑒  𝐻1.008  𝑔  𝐻

=  15.8730  𝑚𝑜𝑙𝑒𝑠  𝐻

  3  

6.9935  𝑚𝑜𝑙𝑒𝑠  𝐶 ∶ 15.8730  𝑚𝑜𝑙𝑒𝑠  𝐻  

1: 2.2698 4: 9

𝐸𝐹 = 𝐶!𝐻!

𝑀𝑜𝑙𝑎𝑟  𝑀𝑎𝑠𝑠  𝐸𝐹 = 57.116  𝑔

𝑋 =114  𝑔/𝑚𝑜𝑙𝑒

57.116  𝑔/𝑚𝑜𝑙𝑒= 2

𝑀𝐹 = 𝐶!𝐻!"

9. You have a 5.40 gram sample that contains 0.88 grams of hydrogen and 4.52 grams of Carbon.

If the molar mass of the compound is 86 grams/mole, what are the empirical and molecular formulas?

4.52  𝑔  𝐶

1  𝑥  

1  𝑚𝑜𝑙𝑒  𝐶12.011  𝑔  𝐶

=  0.3763  𝑚𝑜𝑙𝑒𝑠  𝐶

0.88  𝑔  𝐻

1  𝑥  

1  𝑚𝑜𝑙𝑒  𝐻1.008  𝑔  𝐻

=  0.8730  𝑚𝑜𝑙𝑒𝑠  𝐻

0.3763  𝑚𝑜𝑙𝑒𝑠  𝐶 ∶ 0.8730  𝑚𝑜𝑙𝑒𝑠  𝐻  

1: 2.2319 3: 7

𝐸𝐹 = 𝐶!𝐻!

𝑀𝑜𝑙𝑎𝑟  𝑀𝑎𝑠𝑠  𝐸𝐹 = 43.089  𝑔

𝑋 =86  𝑔/𝑚𝑜𝑙𝑒

43.089  𝑔/𝑚𝑜𝑙𝑒= 2

𝑀𝐹 = 𝐶!𝐻!"

10. Determine the volume of Neon gas if you have 483.2 grams of Neon gas at STP.

483.2  𝑔  𝑁𝑒1

 𝑥  1  𝑚𝑜𝑙𝑒  𝑁𝑒20.180  𝑔  𝑁𝑒

 𝑥  22.4  𝐿  𝑁𝑒1  𝑚𝑜𝑙𝑒  𝑁𝑒

= 536.4  𝐿  𝑁𝑒

11. Determine the number of representative particles of Carbon Monoxide gas if you have 930.2 L

of Carbon Monoxide gas at STP.

930.2  𝐿  𝐶𝑂1

 𝑥  1  𝑚𝑜𝑙𝑒  𝐶𝑂22.4  𝐿  𝐶𝑂

 𝑥  6.02  𝑥  10!"  𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠  𝐶𝑂

1  𝑚𝑜𝑙𝑒  𝐶𝑂= 2.500  𝑥  10!"  𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠  𝐶𝑂

12. Determine the volume of Hydrogen gas at STP if you have 8.38 grams of hydrogen gas.

8.38  𝑔  𝐻!

1  𝑥  

1  𝑚𝑜𝑙𝑒  𝐻!2.016  𝑔  𝐻!

 𝑥  22.4  𝐿  𝐻!1  𝑚𝑜𝑙𝑒  𝐻!

= 93.1  𝐿  𝐻!