Mole Calculation Practice Worksheet featuring six chemistry problems on converting between grams, moles, and molecules for various compounds.

Let’s solve each problem one by one, step by step. We’ll use the mole concept and Avogadro’s number (6.022 × 10²³) where needed.

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Problem 1: How many moles are in 25.0 grams of water?

Water is H₂O.
Molar mass of H₂O = (2 × 1.008) + 16.00 ≈ 18.016 g/mol (we can round to 18.02 g/mol for simplicity).

To find moles:
moles = mass / molar mass
= 25.0 g / 18.02 g/mol ≈ 1.387 moles

✔ Let’s check: 18.02 × 1.387 ≈ 25.0 → correct.

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Problem 2: How many grams are in 4.500 moles of Li₂O?

Li₂O: Lithium oxide
Atomic masses: Li = 6.94, O = 16.00
Molar mass = (2 × 6.94) + 16.00 = 13.88 + 16.00 = 29.88 g/mol

Mass = moles × molar mass
= 4.500 mol × 29.88 g/mol = 134.46 grams

✔ Check: 29.88 × 4.5 = 134.46 → correct.

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Problem 3: How many molecules are in 23.0 moles of oxygen?

Oxygen gas is O₂ (diatomic), but since we’re counting molecules, it doesn’t matter — 1 mole of any substance has 6.022 × 10²³ molecules.

Number of molecules = moles × Avogadro’s number
= 23.0 × (6.022 × 10²³)
= 1.385 × 10²⁵ molecules

✔ Calculation: 23 × 6.022 = 138.506 → so 1.385 × 10²⁵ → correct.

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Problem 4: How many moles are in 3.4 × 10²³ molecules of H₂SO₄?

Use: moles = number of molecules / Avogadro’s number
= (3.4 × 10²³) / (6.022 × 10²³)
≈ 3.4 / 6.022 ≈ 0.5646 moles

✔ Check: 0.5646 × 6.022e23 ≈ 3.4e23 → correct.

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Problem 5: How many molecules are in 25.0 grams of NH₃?

First, find molar mass of NH₃:
N = 14.01, H = 1.008 × 3 = 3.024
Total = 14.01 + 3.024 = 17.034 g/mol

Moles = mass / molar mass = 25.0 / 17.034 ≈ 1.4676 moles

Now, molecules = moles × Avogadro’s number
= 1.4676 × (6.022 × 10²³) ≈ 8.838 × 10²³ molecules

✔ Check: 1.4676 × 6.022 ≈ 8.838 → correct.

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Problem 6: How many grams are in 8.200 × 10²² molecules of N₂I₆?

Wait — N₂I₆? That’s not a real compound. Nitrogen and iodine don’t form N₂I₆. But maybe it’s a typo? Perhaps they meant N₂H₆ or something else? But let’s assume it’s intentional for practice.

Actually, looking again — perhaps it’s N₂I₆ as written. Iodine is I, atomic mass ~126.90. Nitrogen is 14.01.

So molar mass of N₂I₆ = (2 × 14.01) + (6 × 126.90)
= 28.02 + 761.40 = 789.42 g/mol

Now, first find moles from molecules:

moles = (8.200 × 10²²) / (6.022 × 10²³)
= 8.200 / 60.22 ≈ 0.13617 moles

Then, mass = moles × molar mass
= 0.13617 × 789.42 ≈ ?

Calculate:
0.13617 × 789.42
First, 0.1 × 789.42 = 78.942
0.03 × 789.42 = 23.6826
0.006 × 789.42 = 4.73652
0.00017 × 789.42 ≈ 0.1342

Add them:
78.942 + 23.6826 = 102.6246
+4.73652 = 107.36112
+0.1342 ≈ 107.495 grams

But wait — let’s do direct multiplication:

0.13617 × 789.42 = ?

Use calculator-style:
13617 × 78942 = too big — better to compute:

0.13617 × 789.42 = (13617/100000) × 789.42
= (13617 × 789.42) / 100000

Alternatively:
0.13617 × 789.42 ≈ let’s approximate:

0.136 × 789.42 = 0.136 × 789.42
= (0.1 × 789.42) + (0.03 × 789.42) + (0.006 × 789.42)
= 78.942 + 23.6826 + 4.73652 = 107.36112

Now add 0.00017 × 789.42 ≈ 0.134 → total ≈ 107.495

But actually, more accurately:

8.200e22 / 6.022e23 = 0.136167...

0.136167 × 789.42 = let’s compute:

789.42 × 0.136167 = ?

Do 789.42 × 0.13 = 102.6246
789.42 × 0.006 = 4.73652
789.42 × 0.000167 ≈ 789.42 × 0.0001 = 0.078942; ×0.000067≈0.0529 → total ≈ 0.1318

So total ≈ 102.6246 + 4.73652 = 107.36112 + 0.1318 ≈ 107.493 grams

We’ll go with 107.5 grams (rounded to 4 significant figures? Input was 8.200 × 10²² → 4 sig figs, molar mass we calculated with 5, so answer should have 4 sig figs.)

But wait — molar mass of N₂I₆: if we use precise values:

N: 14.007 × 2 = 28.014
I: 126.90447 × 6 = 761.42682
Total = 789.44082 g/mol

moles = 8.200e22 / 6.02214076e23 ≈ 0.136167

mass = 0.136167 × 789.44082 ≈ 107.49 grams

So final answer: 107.5 grams (if rounded to 4 sig figs)

But note: 8.200 has 4 sig figs, Avogadro’s number is exact for this purpose? Usually we treat it as 6.022 × 10²³ with 4 sig figs. So yes, 4 sig figs.

Thus, 107.5 grams

However — is N₂I₆ real? Probably not. Maybe it’s a typo for N₂H₆ (hydrazine derivative?) or perhaps NI₃? But since the problem says N₂I₆, we’ll proceed with that.

Alternatively, maybe it’s N₂I₆ as dinitrogen hexaiodide — which doesn’t exist, but for calculation purposes, we’ll use it.

Final Answer for #6: 107.5 grams

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Now, compiling all answers:

1) 1.387 moles
2) 134.5 grams (wait — earlier I had 134.46, but 4.500 has 4 sig figs, molar mass 29.88 has 4, so 134.5 g? Let me check: 4.500 × 29.88 = 134.46 → rounds to 134.5 g with 4 sig figs? Actually, 134.46 has 5 digits, but since both inputs have 4 sig figs, product should have 4. 134.5 g.

But 4.500 × 29.88:
4.500 has 4 sig figs, 29.88 has 4 → product should be reported with 4 sig figs.

134.46 → rounded to 4 sig figs is 134.5 g.

Similarly, problem 1: 25.0 / 18.02 = 1.387 → 25.0 has 3 sig figs, 18.02 has 4 → so answer should have 3 sig figs: 1.39 moles

Ah! Important — significant figures!

Let me redo with proper sig figs.

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Revised with Significant Figures:

Problem 1:
25.0 g (3 sig figs) / 18.02 g/mol (4 sig figs) → answer has 3 sig figs
25.0 / 18.02 = 1.387... → 1.39 moles

Problem 2:
4.500 mol (4 sig figs) × 29.88 g/mol (4 sig figs) → 134.46 → 134.5 g (4 sig figs)

Problem 3:
23.0 moles (3 sig figs) × 6.022e23 (4 sig figs) → answer has 3 sig figs
23.0 × 6.022e23 = 1.38506e25 → 1.39 × 10²⁵ molecules

Problem 4:
3.4 × 10²³ (2 sig figs!) / 6.022e23 (4 sig figs) → answer has 2 sig figs
3.4 / 6.022 = 0.5646 → 0.56 moles

Problem 5:
25.0 g (3 sig figs) / 17.034 g/mol (5 sig figs) → moles = 1.4676 → 3 sig figs: 1.47 moles
Then 1.47 × 6.022e23 = 8.85234e23 → 8.85 × 10²³ molecules (3 sig figs)

Problem 6:
8.200 × 10²² (4 sig figs) / 6.022e23 (4 sig figs) = 0.136167 → 4 sig figs: 0.1362 moles
Molar mass N₂I₆: if we take N=14.01×2=28.02, I=126.9×6=761.4 → total 789.42 g/mol (5 sig figs? But 126.9 has 4, so 761.4 has 4, 28.02 has 4 → sum 789.42 → but addition: decimal places? 28.02 has 2 decimals, 761.4 has 1 → so sum should have 1 decimal? This is messy.

Typically in such problems, we use standard atomic masses and keep intermediate steps with extra digits, then round final answer.

Given input 8.200e22 has 4 sig figs, and assuming molar mass is known precisely, we'll report 4 sig figs.

Earlier calculation: 0.136167 × 789.42 ≈ 107.49 → 107.5 g (4 sig figs)

Yes.

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Final Answers with Correct Sig Figs:

1) 1.39 moles
2) 134.5 grams
3) 1.39 × 10²⁵ molecules
4) 0.56 moles
5) 8.85 × 10²³ molecules
6) 107.5 grams

Note: For problem 6, if N₂I₆ is not intended, but since it's given, we proceed.

Final Answer:
1) 1.39 moles
2) 134.5 grams
3) 1.39 × 10²⁵ molecules
4) 0.56 moles
5) 8.85 × 10²³ molecules
6) 107.5 grams