Exam Page 1
Show the calculation of the molecular weight for the following two compounds, reporting your
answer to 2 places after the decimal.
Al2(CO3)3
Al2(CO3)3
Al2 = 2 x 26.98 = 53.98
3C = 3 x 12.01 = 36.03
9O = 9 x 16.00 = 144
Ans: 243.01
C8H6NO4Cl
C8H6NO4CI
C = 8 x 12.01 = 96.08
H= 6 x 1.00 = 6
NO = 4 x 259.10 = 1036.4
CI = 35.45 = 35.45
Ans : 1173.93
-4.0 points
Instructor Comments
Calculation error in the first compound.
In the second compound the elements are nitrogen, N and oxygen, O. This is not the
element nobelium, No.
The compound contains one nitrogen atom and 4 oxygen atoms. The first letter of an
element's symbol is always a capital letter. The second, if there is one, is always lower case.
Answer Key
Show the calculation of the molecular weight for the following two compounds, reporting your
answer to 2 places after the decimal.
Al2(CO3)3
2Al + 3C + 9O = 233.99
C8H6NO4Cl
8C + 6H + N + 4O + Cl = 215.59
, Exam Page 2
Show the calculation of the number of moles in the given amount of the following 2
substances.
Report your answer to 3 significant figures.
13.0 grams of Al2(SO4)3
13.0g of Al2(SO4)3
13.0 ÷ 342.17 = 0.037 mol
16.0 grams of C7H5NOBr
16.0 ÷ 428.07 = 0.037 mol
-4.0 points
Instructor Comments
Calculation error in the first compound.
Error in the calculation of molecular weight in the second compound.
Answer Key
Show the calculation of the number of moles in the given amount of the following 2
substances. Report your answer to 3 significant figures.
13.0 grams of Al2(SO4)3
Moles = grams / molecular weight = 13..17 = 0.0380 mole
16.0 grams of C7H5NOBr
Moles = grams / molecular weight = 16..02 = 0.0804 mole
Exam Page 3
Show the calculation of the number of grams in the given amount of the following 2 substances.
Report your answer to 1 place after the decimal.
1.20 moles of Al2(CO3)3
1.20 mol x 233.99 = 280.7g
Show the calculation of the molecular weight for the following two compounds, reporting your
answer to 2 places after the decimal.
Al2(CO3)3
Al2(CO3)3
Al2 = 2 x 26.98 = 53.98
3C = 3 x 12.01 = 36.03
9O = 9 x 16.00 = 144
Ans: 243.01
C8H6NO4Cl
C8H6NO4CI
C = 8 x 12.01 = 96.08
H= 6 x 1.00 = 6
NO = 4 x 259.10 = 1036.4
CI = 35.45 = 35.45
Ans : 1173.93
-4.0 points
Instructor Comments
Calculation error in the first compound.
In the second compound the elements are nitrogen, N and oxygen, O. This is not the
element nobelium, No.
The compound contains one nitrogen atom and 4 oxygen atoms. The first letter of an
element's symbol is always a capital letter. The second, if there is one, is always lower case.
Answer Key
Show the calculation of the molecular weight for the following two compounds, reporting your
answer to 2 places after the decimal.
Al2(CO3)3
2Al + 3C + 9O = 233.99
C8H6NO4Cl
8C + 6H + N + 4O + Cl = 215.59
, Exam Page 2
Show the calculation of the number of moles in the given amount of the following 2
substances.
Report your answer to 3 significant figures.
13.0 grams of Al2(SO4)3
13.0g of Al2(SO4)3
13.0 ÷ 342.17 = 0.037 mol
16.0 grams of C7H5NOBr
16.0 ÷ 428.07 = 0.037 mol
-4.0 points
Instructor Comments
Calculation error in the first compound.
Error in the calculation of molecular weight in the second compound.
Answer Key
Show the calculation of the number of moles in the given amount of the following 2
substances. Report your answer to 3 significant figures.
13.0 grams of Al2(SO4)3
Moles = grams / molecular weight = 13..17 = 0.0380 mole
16.0 grams of C7H5NOBr
Moles = grams / molecular weight = 16..02 = 0.0804 mole
Exam Page 3
Show the calculation of the number of grams in the given amount of the following 2 substances.
Report your answer to 1 place after the decimal.
1.20 moles of Al2(CO3)3
1.20 mol x 233.99 = 280.7g
