Solved Molar Mass Practice Worksheet Find the molar masses | Chegg.com - Free Printable
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Step-by-step solution for: Solved Molar Mass Practice Worksheet Find the molar masses | Chegg.com
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Show Answer Key & Explanations
Step-by-step solution for: Solved Molar Mass Practice Worksheet Find the molar masses | Chegg.com
To find the formula mass of each compound, we add up the atomic masses of all the atoms in the formula. We use atomic masses rounded to the nearest tenth (one decimal place), as instructed.
Here are the standard atomic masses we’ll use (rounded to tenths):
- Ag = 107.9
- N = 14.0
- O = 16.0
- Ni = 58.7
- S = 32.1
- Ca = 40.1
- P = 31.0
- Hg = 200.6
- Fe = 55.8
- K = 39.1
- Br = 79.9
- Be = 9.0
- Cr = 52.0
- Co = 58.9
- Cl = 35.5
- Cu = 63.5
- C = 12.0
- H = 1.0
- Zn = 65.4
- Mg = 24.3
- Al = 27.0
- Na = 23.0
Now let’s calculate each one step by step.
---
1. AgNO₂
Ag: 1 × 107.9 = 107.9
N: 1 × 14.0 = 14.0
O: 2 × 16.0 = 32.0
Total = 107.9 + 14.0 + 32.0 = 153.9
---
2. NiSO₃
Ni: 1 × 58.7 = 58.7
S: 1 × 32.1 = 32.1
O: 3 × 16.0 = 48.0
Total = 58.7 + 32.1 + 48.0 = 138.8
---
3. Ca₃(PO₄)₂
Ca: 3 × 40.1 = 120.3
P: 2 × 31.0 = 62.0
O: 8 × 16.0 = 128.0 (because PO₄ has 4 O, and there are 2 PO₄ → 4×2=8)
Total = 120.3 + 62.0 + 128.0 = 310.3
---
4. HgSO₄
Hg: 1 × 200.6 = 200.6
S: 1 × 32.1 = 32.1
O: 4 × 16.0 = 64.0
Total = 200.6 + 32.1 + 64.0 = 296.7
---
5. Fe(NO₃)₃
Fe: 1 × 55.8 = 55.8
N: 3 × 14.0 = 42.0
O: 9 × 16.0 = 144.0 (each NO₃ has 3 O, 3 groups → 3×3=9)
Total = 55.8 + 42.0 + 144.0 = 241.8
---
6. KBr
K: 1 × 39.1 = 39.1
Br: 1 × 79.9 = 79.9
Total = 39.1 + 79.9 = 119.0
---
7. BeCr₂O₇
Be: 1 × 9.0 = 9.0
Cr: 2 × 52.0 = 104.0
O: 7 × 16.0 = 112.0
Total = 9.0 + 104.0 + 112.0 = 225.0
---
8. Co(ClO₃)₂
Co: 1 × 58.9 = 58.9
Cl: 2 × 35.5 = 71.0
O: 6 × 16.0 = 96.0 (each ClO₃ has 3 O, 2 groups → 3×2=6)
Total = 58.9 + 71.0 + 96.0 = 225.9
---
9. Cu₂C₄H₄O₆
Cu: 2 × 63.5 = 127.0
C: 4 × 12.0 = 48.0
H: 4 × 1.0 = 4.0
O: 6 × 16.0 = 96.0
Total = 127.0 + 48.0 + 4.0 + 96.0 = 275.0
---
10. CuSO₄ · 7 H₂O
First, CuSO₄:
Cu: 63.5
S: 32.1
O: 4 × 16.0 = 64.0 → subtotal = 63.5 + 32.1 + 64.0 = 159.6
Then 7 H₂O:
Each H₂O = 2×1.0 + 16.0 = 18.0
7 × 18.0 = 126.0
Total = 159.6 + 126.0 = 285.6
---
11. ZnCl₂
Zn: 1 × 65.4 = 65.4
Cl: 2 × 35.5 = 71.0
Total = 65.4 + 71.0 = 136.4
---
12. K₃PO₄
K: 3 × 39.1 = 117.3
P: 1 × 31.0 = 31.0
O: 4 × 16.0 = 64.0
Total = 117.3 + 31.0 + 64.0 = 212.3
---
13. Al₂(SO₄)₃
Al: 2 × 27.0 = 54.0
S: 3 × 32.1 = 96.3
O: 12 × 16.0 = 192.0 (each SO₄ has 4 O, 3 groups → 4×3=12)
Total = 54.0 + 96.3 + 192.0 = 342.3
---
14. MgCrO₄
Mg: 1 × 24.3 = 24.3
Cr: 1 × 52.0 = 52.0
O: 4 × 16.0 = 64.0
Total = 24.3 + 52.0 + 64.0 = 140.3
---
15. CaC₄H₄O₆
Ca: 1 × 40.1 = 40.1
C: 4 × 12.0 = 48.0
H: 4 × 1.0 = 4.0
O: 6 × 16.0 = 96.0
Total = 40.1 + 48.0 + 4.0 + 96.0 = 188.1
---
16. NaCl
Na: 1 × 23.0 = 23.0
Cl: 1 × 35.5 = 35.5
Total = 23.0 + 35.5 = 58.5
---
17. K₂Cr₂O₇
K: 2 × 39.1 = 78.2
Cr: 2 × 52.0 = 104.0
O: 7 × 16.0 = 112.0
Total = 78.2 + 104.0 + 112.0 = 294.2
---
18. H₂SO₄
H: 2 × 1.0 = 2.0
S: 1 × 32.1 = 32.1
O: 4 × 16.0 = 64.0
Total = 2.0 + 32.1 + 64.0 = 98.1
---
19. Cu(OH)₂
Cu: 1 × 63.5 = 63.5
O: 2 × 16.0 = 32.0
H: 2 × 1.0 = 2.0
Total = 63.5 + 32.0 + 2.0 = 97.5
---
20. MgSO₄ · 5 H₂O
First, MgSO₄:
Mg: 24.3
S: 32.1
O: 4 × 16.0 = 64.0 → subtotal = 24.3 + 32.1 + 64.0 = 120.4
Then 5 H₂O:
Each H₂O = 18.0
5 × 18.0 = 90.0
Total = 120.4 + 90.0 = 210.4
---
Final Answer:
1. 153.9
2. 138.8
3. 310.3
4. 296.7
5. 241.8
6. 119.0
7. 225.0
8. 225.9
9. 275.0
10. 285.6
11. 136.4
12. 212.3
13. 342.3
14. 140.3
15. 188.1
16. 58.5
17. 294.2
18. 98.1
19. 97.5
20. 210.4
Here are the standard atomic masses we’ll use (rounded to tenths):
- Ag = 107.9
- N = 14.0
- O = 16.0
- Ni = 58.7
- S = 32.1
- Ca = 40.1
- P = 31.0
- Hg = 200.6
- Fe = 55.8
- K = 39.1
- Br = 79.9
- Be = 9.0
- Cr = 52.0
- Co = 58.9
- Cl = 35.5
- Cu = 63.5
- C = 12.0
- H = 1.0
- Zn = 65.4
- Mg = 24.3
- Al = 27.0
- Na = 23.0
Now let’s calculate each one step by step.
---
1. AgNO₂
Ag: 1 × 107.9 = 107.9
N: 1 × 14.0 = 14.0
O: 2 × 16.0 = 32.0
Total = 107.9 + 14.0 + 32.0 = 153.9
---
2. NiSO₃
Ni: 1 × 58.7 = 58.7
S: 1 × 32.1 = 32.1
O: 3 × 16.0 = 48.0
Total = 58.7 + 32.1 + 48.0 = 138.8
---
3. Ca₃(PO₄)₂
Ca: 3 × 40.1 = 120.3
P: 2 × 31.0 = 62.0
O: 8 × 16.0 = 128.0 (because PO₄ has 4 O, and there are 2 PO₄ → 4×2=8)
Total = 120.3 + 62.0 + 128.0 = 310.3
---
4. HgSO₄
Hg: 1 × 200.6 = 200.6
S: 1 × 32.1 = 32.1
O: 4 × 16.0 = 64.0
Total = 200.6 + 32.1 + 64.0 = 296.7
---
5. Fe(NO₃)₃
Fe: 1 × 55.8 = 55.8
N: 3 × 14.0 = 42.0
O: 9 × 16.0 = 144.0 (each NO₃ has 3 O, 3 groups → 3×3=9)
Total = 55.8 + 42.0 + 144.0 = 241.8
---
6. KBr
K: 1 × 39.1 = 39.1
Br: 1 × 79.9 = 79.9
Total = 39.1 + 79.9 = 119.0
---
7. BeCr₂O₇
Be: 1 × 9.0 = 9.0
Cr: 2 × 52.0 = 104.0
O: 7 × 16.0 = 112.0
Total = 9.0 + 104.0 + 112.0 = 225.0
---
8. Co(ClO₃)₂
Co: 1 × 58.9 = 58.9
Cl: 2 × 35.5 = 71.0
O: 6 × 16.0 = 96.0 (each ClO₃ has 3 O, 2 groups → 3×2=6)
Total = 58.9 + 71.0 + 96.0 = 225.9
---
9. Cu₂C₄H₄O₆
Cu: 2 × 63.5 = 127.0
C: 4 × 12.0 = 48.0
H: 4 × 1.0 = 4.0
O: 6 × 16.0 = 96.0
Total = 127.0 + 48.0 + 4.0 + 96.0 = 275.0
---
10. CuSO₄ · 7 H₂O
First, CuSO₄:
Cu: 63.5
S: 32.1
O: 4 × 16.0 = 64.0 → subtotal = 63.5 + 32.1 + 64.0 = 159.6
Then 7 H₂O:
Each H₂O = 2×1.0 + 16.0 = 18.0
7 × 18.0 = 126.0
Total = 159.6 + 126.0 = 285.6
---
11. ZnCl₂
Zn: 1 × 65.4 = 65.4
Cl: 2 × 35.5 = 71.0
Total = 65.4 + 71.0 = 136.4
---
12. K₃PO₄
K: 3 × 39.1 = 117.3
P: 1 × 31.0 = 31.0
O: 4 × 16.0 = 64.0
Total = 117.3 + 31.0 + 64.0 = 212.3
---
13. Al₂(SO₄)₃
Al: 2 × 27.0 = 54.0
S: 3 × 32.1 = 96.3
O: 12 × 16.0 = 192.0 (each SO₄ has 4 O, 3 groups → 4×3=12)
Total = 54.0 + 96.3 + 192.0 = 342.3
---
14. MgCrO₄
Mg: 1 × 24.3 = 24.3
Cr: 1 × 52.0 = 52.0
O: 4 × 16.0 = 64.0
Total = 24.3 + 52.0 + 64.0 = 140.3
---
15. CaC₄H₄O₆
Ca: 1 × 40.1 = 40.1
C: 4 × 12.0 = 48.0
H: 4 × 1.0 = 4.0
O: 6 × 16.0 = 96.0
Total = 40.1 + 48.0 + 4.0 + 96.0 = 188.1
---
16. NaCl
Na: 1 × 23.0 = 23.0
Cl: 1 × 35.5 = 35.5
Total = 23.0 + 35.5 = 58.5
---
17. K₂Cr₂O₇
K: 2 × 39.1 = 78.2
Cr: 2 × 52.0 = 104.0
O: 7 × 16.0 = 112.0
Total = 78.2 + 104.0 + 112.0 = 294.2
---
18. H₂SO₄
H: 2 × 1.0 = 2.0
S: 1 × 32.1 = 32.1
O: 4 × 16.0 = 64.0
Total = 2.0 + 32.1 + 64.0 = 98.1
---
19. Cu(OH)₂
Cu: 1 × 63.5 = 63.5
O: 2 × 16.0 = 32.0
H: 2 × 1.0 = 2.0
Total = 63.5 + 32.0 + 2.0 = 97.5
---
20. MgSO₄ · 5 H₂O
First, MgSO₄:
Mg: 24.3
S: 32.1
O: 4 × 16.0 = 64.0 → subtotal = 24.3 + 32.1 + 64.0 = 120.4
Then 5 H₂O:
Each H₂O = 18.0
5 × 18.0 = 90.0
Total = 120.4 + 90.0 = 210.4
---
Final Answer:
1. 153.9
2. 138.8
3. 310.3
4. 296.7
5. 241.8
6. 119.0
7. 225.0
8. 225.9
9. 275.0
10. 285.6
11. 136.4
12. 212.3
13. 342.3
14. 140.3
15. 188.1
16. 58.5
17. 294.2
18. 98.1
19. 97.5
20. 210.4
Parent Tip: Review the logic above to help your child master the concept of formula mass worksheet.
