Molar Mass Practice Worksheet - Free Printable

To find the molar mass of each compound, we add up the atomic masses of all the atoms in the formula. We use the periodic table to find the atomic mass of each element (rounded to two decimal places).

Here are the approximate atomic masses used:
* Na: 22.99 g/mol
* Br: 79.90 g/mol
* Pb: 207.20 g/mol
* S: 32.06 g/mol
* O: 16.00 g/mol
* Ca: 40.08 g/mol
* H: 1.01 g/mol
* P: 30.97 g/mol
* N: 14.01 g/mol
* C: 12.01 g/mol
* Fe: 55.85 g/mol
* Zn: 65.38 g/mol
* Ag: 107.87 g/mol
* F: 19.00 g/mol

Step-by-Step Calculations:

1) NaBr
* Na: $1 \times 22.99 = 22.99$
* Br: $1 \times 79.90 = 79.90$
* Total: $22.99 + 79.90 = 102.89$ g/mol

2) PbSO₄
* Pb: $1 \times 207.20 = 207.20$
* S: $1 \times 32.06 = 32.06$
* O: $4 \times 16.00 = 64.00$
* Total: $207.20 + 32.06 + 64.00 = 303.26$ g/mol

3) Ca(OH)₂
* Ca: $1 \times 40.08 = 40.08$
* O: $2 \times 16.00 = 32.00$ (Note: The subscript 2 outside the parenthesis applies to both O and H)
* H: $2 \times 1.01 = 2.02$
* Total: $40.08 + 32.00 + 2.02 = 74.10$ g/mol

4) Na₃PO₄
* Na: $3 \times 22.99 = 68.97$
* P: $1 \times 30.97 = 30.97$
* O: $4 \times 16.00 = 64.00$
* Total: $68.97 + 30.97 + 64.00 = 163.94$ g/mol

5) (NH₄)₂CO₃
* N: $2 \times 14.01 = 28.02$
* H: $8 \times 1.01 = 8.08$ ($4 \times 2$ hydrogens)
* C: $1 \times 12.01 = 12.01$
* O: $3 \times 16.00 = 48.00$
* Total: $28.02 + 8.08 + 12.01 + 48.00 = 96.11$ g/mol

6) C₆H₁₂O₆
* C: $6 \times 12.01 = 72.06$
* H: $12 \times 1.01 = 12.12$
* O: $6 \times 16.00 = 96.00$
* Total: $72.06 + 12.12 + 96.00 = 180.18$ g/mol

7) Fe₃(PO₄)₂
* Fe: $3 \times 55.85 = 167.55$
* P: $2 \times 30.97 = 61.94$
* O: $8 \times 16.00 = 128.00$ ($4 \times 2$ oxygens)
* Total: $167.55 + 61.94 + 128.00 = 357.49$ (or 357.48 depending on rounding of Iron) -> Let's stick to standard rounding: $167.55 + 61.94 + 128.00 = 357.49$ g/mol. *Self-correction: Using precise values often yields 357.48. Let's re-calculate: $3(55.845) + 2(30.974) + 8(15.999) = 167.535 + 61.948 + 127.992 = 357.475$. Rounding to two decimals gives 357.48.*

8) (NH₄)₂S
* N: $2 \times 14.01 = 28.02$
* H: $8 \times 1.01 = 8.08$
* S: $1 \times 32.06 = 32.06$
* Total: $28.02 + 8.08 + 32.06 = 68.16$ (or 68.14/68.15 depending on H precision). Let's use standard school values: $N=14.01, H=1.008, S=32.06$.
* $N: 28.02$
* $H: 8 \times 1.008 = 8.064$
* $S: 32.06$
* Total: $68.144 \rightarrow 68.14$ g/mol.

9) Zn(C₂H₃O₂)₂
* Zn: $1 \times 65.38 = 65.38$
* C: $4 \times 12.01 = 48.04$ ($2 \times 2$ carbons)
* H: $6 \times 1.01 = 6.06$ ($3 \times 2$ hydrogens)
* O: $4 \times 16.00 = 64.00$ ($2 \times 2$ oxygens)
* Total: $65.38 + 48.04 + 6.06 + 64.00 = 183.48$ g/mol

10) AgF
* Ag: $1 \times 107.87 = 107.87$
* F: $1 \times 19.00 = 19.00$
* Total: $107.87 + 19.00 = 126.87$ g/mol

Final Answer:
1) 102.89 g/mol
2) 303.26 g/mol
3) 74.10 g/mol
4) 163.94 g/mol
5) 96.11 g/mol
6) 180.18 g/mol
7) 357.48 g/mol
8) 68.14 g/mol
9) 183.48 g/mol
10) 126.87 g/mol