Question

Calculate the molecular mass of the following molecules:
a) CaCO
3



b) H
2

SO
4



c) NH
3



d) Fe(NO
3

)
3



e) C
6

H
12

O
6



f) C
6

H
14



g) HCl

h) NaOH

i) C
2

H
2



j) Al(OH)
3



k) M (M is unspecified; please clarify the molecule)

l) CO
2



m) HNO
3



n) CuSO
4



o) Fe
2

O
3

(corrected from Fe
2

O
2

, which is not a common compound)

Answers

GPT-4o mini
To calculate the molecular mass of the given molecules, we sum up the atomic masses of each element in the molecule. Below are the atomic masses of the relevant elements:

- Carbon (C): ~12.01 g/mol
- Calcium (Ca): ~40.08 g/mol
- Oxygen (O): ~16.00 g/mol
- Hydrogen (H): ~1.01 g/mol
- Sulfur (S): ~32.07 g/mol
- Nitrogen (N): ~14.01 g/mol
- Iron (Fe): ~55.85 g/mol
- Copper (Cu): ~63.55 g/mol
- Sodium (Na): ~22.99 g/mol
- Aluminum (Al): ~26.98 g/mol

Now, let's conduct the calculations for each molecule:

a) **CaCO₃**
- Ca: 1 × 40.08 = 40.08
- C: 1 × 12.01 = 12.01
- O: 3 × 16.00 = 48.00
- Total = 40.08 + 12.01 + 48.00 = **100.09 g/mol**

b) **H₂SO₄**
- H: 2 × 1.01 = 2.02
- S: 1 × 32.07 = 32.07
- O: 4 × 16.00 = 64.00
- Total = 2.02 + 32.07 + 64.00 = **98.09 g/mol**

c) **NH₃**
- N: 1 × 14.01 = 14.01
- H: 3 × 1.01 = 3.03
- Total = 14.01 + 3.03 = **17.04 g/mol**

d) **Fe(NO₃)₃**
- Fe: 1 × 55.85 = 55.85
- N: 3 × 14.01 = 42.03
- O: 9 × 16.00 = 144.00
- Total = 55.85 + 42.03 + 144.00 = **241.88 g/mol**

e) **C₆H₁₂O₆**
- C: 6 × 12.01 = 72.06
- H: 12 × 1.01 = 12.12
- O: 6 × 16.00 = 96.00
- Total = 72.06 + 12.12 + 96.00 = **180.18 g/mol**

f) **C₆H₁₄**
- C: 6 × 12.01 = 72.06
- H: 14 × 1.01 = 14.14
- Total = 72.06 + 14.14 = **86.20 g/mol**

g) **HCl**
- H: 1 × 1.01 = 1.01
- Cl: 1 × 35.45 = 35.45
- Total = 1.01 + 35.45 = **36.46 g/mol**

h) **NaOH**
- Na: 1 × 22.99 = 22.99
- O: 1 × 16.00 = 16.00
- H: 1 × 1.01 = 1.01
- Total = 22.99 + 16.00 + 1.01 = **40.00 g/mol**

i) **C₂H₂**
- C: 2 × 12.01 = 24.02
- H: 2 × 1.01 = 2.02
- Total = 24.02 + 2.02 = **26.04 g/mol**

j) **Al(OH)₃**
- Al: 1 × 26.98 = 26.98
- O: 3 × 16.00 = 48.00
- H: 3 × 1.01 = 3.03
- Total = 26.98 + 48.00 + 3.03 = **78.01 g/mol**

k) **M** (As you haven’t specified the molecule, we cannot calculate a molecular mass.)

l) **CO₂**
- C: 1 × 12.01 = 12.01
- O: 2 × 16.00 = 32.00
- Total = 12.01 + 32.00 = **44.01 g/mol**

m) **HNO₃**
- H: 1 × 1.01 = 1.01
- N: 1 × 14.01 = 14.01
- O: 3 × 16.00 = 48.00
- Total = 1.01 + 14.01 + 48.00 = **63.02 g/mol**

n) **CuSO₄**
- Cu: 1 × 63.55 = 63.55
- S: 1 × 32.07 = 32.07
- O: 4 × 16.00 = 64.00
- Total = 63.55 + 32.07 + 64.00 = **159.62 g/mol**

o) **Fe₂O₃**
- Fe: 2 × 55.85 = 111.70
- O: 3 × 16.00 = 48.00
- Total = 111.70 + 48.00 = **159.70 g/mol**

If you need molecular mass for any additional specified compounds, feel free to ask!