Molar Mass and Mole Calculations Worksheet with detailed problems and solutions.

Let’s go through each problem step by step. I’ll show you how to solve them so you can understand the process — not just copy answers.

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Problem 1: Calculate molar mass for each compound

Molar mass = sum of atomic masses of all atoms in the formula (use periodic table values).

a. NaCl
Na = 22.99, Cl = 35.45 → 22.99 + 35.45 = 58.44 g/mol

b. Fe₂O₃
Fe = 55.85 × 2 = 111.70
O = 16.00 × 3 = 48.00
Total = 111.70 + 48.00 = 159.70 g/mol

c. MgCO₃
Mg = 24.31, C = 12.01, O = 16.00 × 3 = 48.00
Total = 24.31 + 12.01 + 48.00 = 84.32 g/mol

d. Ba(OH)₂
Ba = 137.33
O = 16.00 × 2 = 32.00
H = 1.01 × 2 = 2.02
Total = 137.33 + 32.00 + 2.02 = 171.35 g/mol

e. (NH₄)₃PO₄
N = 14.01 × 3 = 42.03
H = 1.01 × 12 = 12.12 (because NH₄ has 4 H, and there are 3 groups → 4×3=12)
P = 30.97
O = 16.00 × 4 = 64.00
Total = 42.03 + 12.12 + 30.97 + 64.00 = 149.12 g/mol

f. Ca₃(PO₄)₂
Ca = 40.08 × 3 = 120.24
P = 30.97 × 2 = 61.94
O = 16.00 × 8 = 128.00 (because PO₄ has 4 O, and there are 2 groups → 4×2=8)
Total = 120.24 + 61.94 + 128.00 = 310.18 g/mol

g. C₆H₁₂O₆
C = 12.01 × 6 = 72.06
H = 1.01 × 12 = 12.12
O = 16.00 × 6 = 96.00
Total = 72.06 + 12.12 + 96.00 = 180.18 g/mol

h. C₅H₅N
C = 12.01 × 5 = 60.05
H = 1.01 × 5 = 5.05
N = 14.01
Total = 60.05 + 5.05 + 14.01 = 79.11 g/mol

✔ All match the given answers.

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Problem 2: Calculate moles from mass

Formula: moles = mass (g) ÷ molar mass (g/mol)

a. 45.0 g acetic acid, CH₃COOH

First, find molar mass of CH₃COOH:
C₂H₄O₂ → C=12.01×2=24.02, H=1.01×4=4.04, O=16.00×2=32.00
Total = 24.02 + 4.04 + 32.00 = 60.06 g/mol

Now: moles = 45.0 ÷ 60.06 ≈ 0.749 mol → rounds to 0.75 mol

b. 7.04 g lead(II) nitrate, Pb(NO₃)₂

Molar mass:
Pb = 207.2
N = 14.01 × 2 = 28.02
O = 16.00 × 6 = 96.00
Total = 207.2 + 28.02 + 96.00 = 331.22 g/mol

moles = 7.04 ÷ 331.22 ≈ 0.02125 → rounds to 0.02 mol

✔ Matches given answers.

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Problem 3: Calculate mass from moles

Formula: mass = moles × molar mass

a. 3.00 mol SeOBr₂

Molar mass:
Se = 78.97
O = 16.00
Br = 79.90 × 2 = 159.80
Total = 78.97 + 16.00 + 159.80 = 254.77 g/mol

mass = 3.00 × 254.77 = 764.31 g

b. 488 mol CaCO₃

Molar mass:
Ca = 40.08
C = 12.01
O = 16.00 × 3 = 48.00
Total = 40.08 + 12.01 + 48.00 = 100.09 g/mol

mass = 488 × 100.09 = let’s calculate:

488 × 100 = 48,800
488 × 0.09 = 43.92
Total = 48,800 + 43.92 = 48,843.92 g

✔ Matches given answers.

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Problem 4: Calculate number of molecules or formula units

Use Avogadro’s number: 6.02 × 10²³ particles per mole

Formula: particles = moles × 6.02 × 10²³

a. 4.27 mol WO₃

particles = 4.27 × 6.02 × 10²³
First: 4.27 × 6.02 = 25.7054
So: 2.57 × 10²⁴ (rounded to 3 sig figs)

b. 0.989 mol C₆H₅NO₂

particles = 0.989 × 6.02 × 10²³
0.989 × 6.02 = 5.95378
So: 5.95 × 10²³

✔ Matches given answers.

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Problem 5: Calculate mass from number of molecules/formula units

Two steps:
1. Convert particles → moles: moles = particles ÷ 6.02 × 10²³
2. Then: mass = moles × molar mass

a. 8.39 × 10²³ molecules of fluorine (F₂)

Note: Fluorine gas is F₂, so molar mass = 2 × 19.00 = 38.00 g/mol

Step 1: moles = 8.39 × 10²³ ÷ 6.02 × 10²³ = 8.39 ÷ 6.02 ≈ 1.3937 mol

Step 2: mass = 1.3937 × 38.00 ≈ 52.96 g

b. 6.82 × 10²⁴ formula units of beryllium sulfate (BeSO₄)

Wait — actually, beryllium sulfate is BeSO₄? Let me check.

Actually, beryllium is Be²⁺, sulfate is SO₄²⁻ → so yes, BeSO₄

But wait — the worksheet says “beryllium sulfate” and gives calculation with M = 9.01 + 32.06 + 4(16) = 105.07 → that’s BeSO₄.

But then it multiplies by 11.33? That doesn’t make sense unless...

Wait — looking at their work:

They did:
M = 9.01 + 32.06 + 4(16) = 105.07 → correct for BeSO₄
Then: 6.82 × 10²⁴ / 6.02 × 10²³ = 11.33 mol
Then: 105.07 × 11.33 = ?

Let me compute:

6.82e24 ÷ 6.02e23 = 68.2 ÷ 6.02 ≈ 11.3289 → ~11.33 mol

Then mass = 11.33 × 105.07

Calculate:

11 × 105.07 = 1155.77
0.33 × 105.07 ≈ 34.6731
Total ≈ 1155.77 + 34.67 = 1190.44 g

But they have 1,190.33 g — probably rounding difference.

Let me do exact:

11.33 × 105.07 = ?

Break it down:

10 × 105.07 = 1050.7
1.33 × 105.07 = ?

1 × 105.07 = 105.07
0.33 × 105.07 = 34.6731
So 105.07 + 34.6731 = 139.7431

Total = 1050.7 + 139.7431 = 1190.4431 g

Hmm — but they got 1,190.33. Maybe they used slightly different atomic masses?

In their calc: M = 9.01 + 32.06 + 64 = 105.07 — same as mine.

And 6.82e24 / 6.02e23 = 11.3289... which they rounded to 11.33

Then 105.07 × 11.33:

Let me multiply exactly:

105.07 × 11.33

= 105.07 × (11 + 0.33)
= 105.07×11 = 1155.77
105.07×0.33 = 34.6731
Sum = 1190.4431 → still 1190.44

But they wrote 1,190.33 — maybe typo? Or perhaps they used 6.022 instead of 6.02?

Try with 6.022:

6.82e24 / 6.022e23 = 11.325...

11.325 × 105.07 ≈ ?

11.325 × 100 = 1132.5
11.325 × 5.07 ≈ 57.41775
Total ≈ 1189.91775 — still not 1190.33

Alternatively, maybe they meant Be₂(SO₄)₃? No, that would be for Al³⁺ etc.

Wait — beryllium sulfate is indeed BeSO₄. But let’s check the answer they gave: 1,190.33 g

Perhaps they calculated:

M = 9.01 + 32.06 + 64 = 105.07
moles = 6.82e24 / 6.02e23 = 11.3289
then 105.07 × 11.3289 = ?

Compute:

105.07 × 11.3289

First, 105.07 × 11 = 1155.77
105.07 × 0.3289 ≈ ?

105.07 × 0.3 = 31.521
105.07 × 0.0289 ≈ 3.036523
Sum ≈ 34.5575

Total ≈ 1155.77 + 34.5575 = 1190.3275 → 1190.33 g

Ah! So they kept more decimals in intermediate step.

So if we use moles = 6.82e24 / 6.02e23 = 11.32890365...

Then 105.07 × 11.32890365 ≈ 1190.33 g

Yes — so it matches when using precise division.

So final answer is 1,190.33 g

✔ All problems solved correctly.

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Final Answer:
All calculations verified and match the provided answers on the worksheet. The solutions involve applying molar mass formulas, mole-mass conversions, and Avogadro's number appropriately for each part.