Ninth grade Lesson Converting Between Masses and Moles - Free Printable

Let’s go through each problem step by step to find the number of moles.

We use this formula:

> moles = mass (g) ÷ molar mass (g/mol)

We’ll calculate the molar mass for each compound using atomic masses, then divide the given mass by that molar mass.

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1. 25 g of NaCl

Atomic masses:
- Na = 22.990
- Cl = 35.453
→ Molar mass of NaCl = 22.990 + 35.453 = 58.443 g/mol

Moles = 25 ÷ 58.443 ≈ 0.4278 mol

Rounded to two decimal places: 0.43 mol

*(Note: The student wrote 0.42 — let’s check their math: 25 ÷ 58.443 = ? Let me recalculate: 58.443 × 0.42 = 24.546 → too low. 58.443 × 0.43 = 25.13 → closer. Actually, 25 ÷ 58.443 = 0.4278 → so 0.43 is better rounded. But maybe they used slightly different atomic masses? Let’s stick with precise calculation.)*

Actually, let’s use more precise division:

25 ÷ 58.443 = 0.42776... → rounds to 0.43 mol if we round to two decimals.

But in the image, they got 0.42 — perhaps they truncated or used rounding earlier. We’ll go with accurate value.

Wait — let’s double-check all calculations carefully as instructed.

Let me do exact divisions:

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Problem 1: 25 g NaCl

Molar mass = 22.990 + 35.453 = 58.443 g/mol

25 ÷ 58.443 = ?

Calculate: 58.443 × 0.427 = 58.443 × 0.4 = 23.3772; 58.443 × 0.027 = ~1.578 → total 24.955 → close to 25.

So 0.4278 → 0.428 mol if three decimals, but typically we report based on significant figures.

Given mass is 25 g → two significant figures.

Molar mass has 5 sig figs, but limiting is 25 → two sig figs.

So answer should be 0.43 mol (since 0.4278 rounds up).

But wait — 25 could be considered exact? In many school problems, they treat it as having two sig figs.

Looking at other answers in image: they have 0.42, 1.27, etc. — seems like they’re keeping two or three decimals.

Let’s proceed consistently and compute all accurately, then round appropriately.

I’ll compute each one precisely and then decide rounding at end.

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Problem 2: 125 g H₂SO₄

Atomic masses:
- H = 1.008 × 2 = 2.016
- S = 32.066
- O = 15.999 × 4 = 63.996
→ Total = 2.016 + 32.066 + 63.996 = 98.078 g/mol

Moles = 125 ÷ 98.078 ≈ ?

98.078 × 1.274 = let’s check: 98.078 × 1.2 = 117.6936; 98.078 × 0.074 = ~7.257 → total 124.95 → very close to 125.

So 125 ÷ 98.078 = 1.2745... → 1.27 mol (if rounded to three sig figs) or 1.275? Wait, 125 has three sig figs, molar mass has five → so answer can have three.

1.2745 → rounds to 1.27 mol? No — 1.2745 to three sig figs is 1.27? Wait, 1.27 has three sig figs, but 1.2745 is closer to 1.27 than 1.28? Actually, 1.2745 — the fourth digit is 4, so round down → 1.27.

But let’s calculate exactly:

125 ÷ 98.078 = 1.274507... → so 1.27 mol if rounded to three significant figures.

In image, they have 1.274 — which is four digits. Maybe they kept more precision.

For consistency, I’ll compute all and then see what makes sense.

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Problem 3: 100. g KMnO₄

Note: “100.” means three significant figures.

Atomic masses:
- K = 39.098
- Mn = 54.938
- O = 15.999 × 4 = 63.996
→ Total = 39.098 + 54.938 + 63.996 = 158.032 g/mol

Moles = 100. ÷ 158.032 ≈ ?

158.032 × 0.6328 = let’s not guess — calculate:

100 ÷ 158.032 = 0.63278... → 0.633 mol (three sig figs)

Image shows 0.63 — probably rounded to two decimals, but since 100. has three sig figs, we should keep three.

0.6328 → rounds to 0.633 mol

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Problem 4: 74 g KCl

Atomic masses:
- K = 39.098
- Cl = 35.453
→ Molar mass = 39.098 + 35.453 = 74.551 g/mol

Moles = 74 ÷ 74.551 ≈ ?

74.551 × 0.9926 = approximately 74.

Calculate: 74 ÷ 74.551 = 0.9926... → 0.993 mol (three sig figs? 74 has two sig figs!)

Ah! Important: 74 g — no decimal, so likely two significant figures.

So answer should have two sig figs.

0.9926 → rounds to 0.99 mol

Image shows 0.99 — correct.

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Problem 5: 35 g CuSO₄

Atomic masses:
- Cu = 63.546
- S = 32.066
- O = 15.999 × 4 = 63.996
→ Total = 63.546 + 32.066 + 63.996 = 159.608 g/mol

Moles = 35 ÷ 159.608 ≈ ?

159.608 × 0.2193 = let’s compute:

35 ÷ 159.608 = 0.21928... → 0.219 mol

Now, 35 g — two sig figs → so answer should be 0.22 mol

Image shows blank, but they started writing .22 — so probably 0.22 mol.

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Now, let’s summarize with proper rounding based on significant figures:

1. 25 g NaCl → 25 has two sig figs → 25 / 58.443 = 0.4278 → 0.43 mol
- But wait: 0.43 has two sig figs? Yes, leading zero doesn’t count, so 4 and 3 are significant → two sig figs. Correct.

2. 125 g H₂SO₄ → 125 has three sig figs → 125 / 98.078 = 1.2745 → 1.27 mol (three sig figs)
- 1.27 has three sig figs.

3. 100. g KMnO₄ → 100. has three sig figs → 100. / 158.032 = 0.6328 → 0.633 mol

4. 74 g KCl → 74 has two sig figs → 74 / 74.551 = 0.9926 → 0.99 mol

5. 35 g CuSO₄ → 35 has two sig figs → 35 / 159.608 = 0.2193 → 0.22 mol

But looking back at the image, for problem 1, they got 0.42 — which might be from using approximate atomic masses or truncating.

To match typical textbook approach, sometimes they use rounded atomic masses.

Let me try with common rounded values often used in schools:

Common approximations:
- Na = 23, Cl = 35.5 → NaCl = 58.5 → 25/58.5 ≈ 0.427 → still 0.43
- Or if they use Na=23, Cl=35.45 → 58.45 → 25/58.45≈0.4277→0.43

Perhaps they did 25 / 58.443 = let's calculate numerically:

58.443 * 0.42 = 24.54606 — less than 25
58.443 * 0.43 = 25.13049 — over 25
Difference: 25 - 24.546 = 0.454; increment per 0.01 is 0.58443, so 0.454 / 0.58443 ≈ 0.777, so 0.42 + 0.00777 = 0.42777 — same as before.

So 0.4278 — if they rounded to two decimal places, it's 0.43, but they wrote 0.42 — maybe a mistake.

Similarly, for problem 2: 125 / 98.078 = 1.2745 — they have 1.274, which is fine if keeping three decimals.

For problem 3: 100 / 158.032 = 0.6328 — they have 0.63, which is rounded to two decimals.

For problem 4: 74 / 74.551 = 0.9926 — they have 0.99, good.

For problem 5: 35 / 159.608 = 0.2193 — should be 0.22 if rounded to two decimals or two sig figs.

Since the instruction is to solve accurately, and verify, I'll provide the most accurate values with appropriate rounding.

But to be consistent with how the student did it (they seem to round to two or three decimals), and since the problems don't specify sig fig rules strictly, I'll give answers as calculated and rounded to three significant figures where possible, or as per input.

Let me list final answers with reasoning:

1. 25 g NaCl: molar mass 58.443 g/mol → 25 / 58.443 = 0.4278 mol → 0.428 mol (but typically reported as 0.43 mol for two sig figs). However, since 25 might be exact, but usually not. I think safest is to use the calculation as is.

Looking at standard practice in such worksheets, they often expect the division result as shown.

Perhaps I should just compute each one as per the numbers given and box the answer.

Let me do precise calculations:

1. 25 / 58.443 = 0.42776 → 0.428 mol (if three decimals) or 0.43 mol

But in the image, for #1, they have 0.42 — which is incorrect; it should be 0.43 if rounded properly.

For accuracy, I'll use calculator-like precision.

Assume we use the atomic masses as given in the image for consistency.

In image:

#1: Na=22.990, Cl=35.453 → sum 58.443 → 25/58.443 = let's calculate: 25 ÷ 58.443 = 0.42776 → so 0.428 if rounded to three decimals, but they wrote 0.42 — perhaps typo or early rounding.

Similarly, #2: H2SO4: H=1.008*2=2.016, S=32.066, O=15.999*4=63.996 → sum 98.078 → 125/98.078 = 1.2745 → they have 1.274 — good.

#3: KMnO4: K=39.098, Mn=54.938, O=15.999*4=63.996 → sum 158.032 → 100/158.032 = 0.63278 → they have 0.63 — rounded down.

#4: KCl: K=39.098, Cl=35.453 → sum 74.551 → 74/74.551 = 0.9926 → they have 0.99 — good.

#5: CuSO4: Cu=63.546, S=32.066, O=63.996 → sum 159.608 → 35/159.608 = 0.21928 → should be 0.219 or 0.22.

For the final answer, since the user wants the correct solution, I'll provide the accurate values.

But to match the format, and since the student has partial answers, I'll complete them correctly.

Final decision: Calculate each with given atomic masses and report to three significant figures or as appropriate.

1. 25 g NaCl: 25 / 58.443 = 0.4278 → 0.428 mol (but perhaps 0.43 is expected)

I recall that in many textbooks, for 25 g NaCl, they say 0.43 mol.

Let's check online or standard value: molar mass NaCl is often taken as 58.44 g/mol, 25/58.44 = 0.4278 -> 0.43 mol.

Similarly, for #5, 35/159.61 = 0.2193 -> 0.22 mol.

So I'll go with:

1. 0.43 mol
2. 1.27 mol
3. 0.633 mol
4. 0.99 mol
5. 0.22 mol

But for #3, 100. / 158.032 = 0.6328 -> 0.633 mol (three sig figs)

For #2, 125 / 98.078 = 1.2745 -> 1.27 mol (three sig figs) — 1.27 has three, yes.

1.2745 rounded to three sig figs is 1.27, since the third digit is 7, fourth is 4<5, so 1.27.

1.27 has three sig figs.

Similarly, 0.6328 to three sig figs is 0.633.

0.9926 to two sig figs is 0.99.

0.2193 to two sig figs is 0.22.

0.4278 to two sig figs is 0.43.

Perfect.

So final answers:

1. 0.43 mol
2. 1.27 mol
3. 0.633 mol
4. 0.99 mol
5. 0.22 mol

Now, for the response, since ExplainAnswer is false, TeachTopic is false, GeneratePractice is false, I only need to provide Final Answer.

But the instruction says: "Complete ALL reasoning and explanations BEFORE writing the Final Answer." and "If an option is false, completely omit that section."

And "Final Answer: [Provide only the final correct answer]"

But it also says "work through the problem step by step" in the critical rule, and "verify your answer before writing it".

So I have done the reasoning above, now I write the final answer.

The final answer should be the list of moles for each.

How to present? Probably as a list.

Since the original has 5 items, I'll list them.

Final Answer:
1. 0.43 mol
2. 1.27 mol
3. 0.633 mol
4. 0.99 mol
5. 0.22 mol

To be precise, let's confirm the calculations once more with exact arithmetic.

Use calculator simulation:

1. 25 / 58.443 = 0.42776 → rounds to 0.43 (two sig figs)
2. 125 / 98.078 = 1.2745 → rounds to 1.27 (three sig figs)
3. 100. / 158.032 = 0.63278 → rounds to 0.633 (three sig figs)
4. 74 / 74.551 = 0.9926 → rounds to 0.99 (two sig figs)
5. 35 / 159.608 = 0.21928 → rounds to 0.22 (two sig figs)

Yes.

Some might argue for #2 to be 1.275, but 1.2745 is closer to 1.27 than 1.28? No, 1.2745 - the digit after 7 is 4, so when rounding to three sig figs, look at the fourth digit: 1.2745, the third sig fig is 7, fourth is 4, so round down, keep 7, so 1.27.

Standard rounding: 1.2745 to three significant figures: the number is between 1.27 and 1.28. The fourth digit is 4, which is less than 5, so we do not round up, so 1.27.

Similarly, 0.63278 to three sig figs: 0.633 because the third digit is 2, fourth is 7>5, so round up 2 to 3? Let's see: 0.63278

Significant figures for 0.63278: first non-zero is 6, so sig figs are 6,3,2,7,8.

To three sig figs: look at the fourth digit, which is 7 > 5, so round up the third digit from 2 to 3, so 0.633.

Yes.

0.9926 to two sig figs: first two digits are 9 and 9, third is 2<5, so 0.99.

0.21928 to two sig figs: first two are 2 and 1, third is 9>5, so round up 1 to 2, so 0.22.

0.4278 to two sig figs: first two are 4 and 2, third is 7>5, so round up 2 to 3, so 0.43.

All correct.

Final Answer:
1. 0.43 mol
2. 1.27 mol
3. 0.633 mol
4. 0.99 mol
5. 0.22 mol