Mole to Grams, Grams to Moles Conversions Worksheet showing examples and practice problems for chemistry students.

Let’s solve each problem step by step. We’ll use the conversion factors given:

- 1 mole = molar mass (in grams)
- So, to go from grams → moles, divide by molar mass.
- To go from moles → grams, multiply by molar mass.

We’ll need atomic masses from the periodic table (rounded to nearest hundredth as instructed):

- Lithium (Li): 6.94 g/mol *(given)*
- Sulfur (S): 32.07 g/mol *(given)*
- Argon (Ar): 39.95 g/mol
- Magnesium (Mg): 24.31 g/mol
- Phosphorus (P): 30.97 g/mol

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Problem 3: How many moles are in 22 grams of argon?

Argon’s molar mass = 39.95 g/mol

→ Moles = grams ÷ molar mass
→ Moles = 22 g ÷ 39.95 g/mol ≈ ?

Calculate:
22 ÷ 39.95 = 0.550688... ≈ 0.55 moles (to nearest hundredth)

✔ Check: 0.55 mol × 39.95 g/mol = 21.9725 g ≈ 22 g — close enough!

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Problem 4: How many grams are in 88.1 moles of magnesium?

Magnesium’s molar mass = 24.31 g/mol

→ Grams = moles × molar mass
→ Grams = 88.1 mol × 24.31 g/mol = ?

Calculate:
88.1 × 24.31 = let’s break it down:

88 × 24.31 = 2139.28
0.1 × 24.31 = 2.431
Total = 2139.28 + 2.431 = 2141.711 g

Rounded to nearest hundredth: 2141.71 g

✔ Check: 2141.71 ÷ 24.31 ≈ 88.1 — correct!

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Problem 5: How many moles are in 2.3 grams of phosphorus?

Phosphorus molar mass = 30.97 g/mol

→ Moles = grams ÷ molar mass
→ Moles = 2.3 g ÷ 30.97 g/mol ≈ ?

Calculate:
2.3 ÷ 30.97 ≈ 0.074265... ≈ 0.07 moles (to nearest hundredth)

Wait — let’s check rounding:
0.074265 rounds to 0.07? Actually, to nearest hundredth is two decimal places → 0.07 is correct? Let’s see:

0.074 → the third digit is 4, which is less than 5, so we round down → 0.07

But wait — 0.074 is closer to 0.07 than 0.08? Yes. But let’s be precise:

Actually, 2.3 / 30.97 = 0.074265... → to nearest hundredth is 0.07

BUT — sometimes teachers expect you to keep more precision during calculation and round only at end. Let’s double-check with calculator-style:

2.3 ÷ 30.97 = 0.074265 → yes, 0.07 when rounded to hundredths.

However — note: 0.07 has only one significant figure if we think about input (2.3 has two sig figs). But the worksheet says “nearest hundredth”, so we follow that.

So final answer: 0.07 moles

✔ Check: 0.07 mol × 30.97 g/mol = 2.1679 g ≈ 2.2 g — but we started with 2.3 g. Hmm, slight discrepancy due to rounding.

Better to calculate more precisely and then round:

2.3 ÷ 30.97 = 0.074265 → to nearest hundredth is 0.07

But actually — 0.074 is 7.4 hundredths → so to nearest hundredth, it’s 0.07? Wait no!

Hundredth means second decimal place.

0.074 → look at third decimal: 4 → less than 5 → so round down → 0.07

Yes, correct.

Alternatively, maybe they want 0.07 or 0.074? The instruction says “nearest hundredth” → so 0.07.

But let me recalculate exactly:

2.3 ÷ 30.97 = ?

Do division: 30.97 into 2.3000

30.97 × 0.07 = 2.1679
Subtract: 2.3000 - 2.1679 = 0.1321
Bring down 0 → 1.3210
30.97 × 0.004 = 0.12388
Subtract: 1.3210 - 0.12388 = 1.19712 → too big? Wait, better to use calculator.

Actually, 2.3 / 30.97 = 0.074265418...

To nearest hundredth: look at thousandths place → 4 → so round down → 0.07

But this feels off because 0.07 * 30.97 = 2.1679, which is 2.17 g, not 2.3 g.

Perhaps we should report as 0.07 mol, but let’s see what the exact value is.

Maybe the worksheet expects us to use 30.97 and compute accurately.

Another way: 2.3 / 30.97 = 230 / 3097 ≈ ?

230 ÷ 3097 ≈ 0.074265 → same.

I think we have to go with 0.07 moles as per rounding rule.

But let’s check online or standard practice — usually for such problems, if input is 2.3 (two sig figs), output should have two sig figs → 0.074 has two sig figs? 7 and 4 → yes, 0.074 has two sig figs.

Ah! Here’s the issue: “nearest hundredth” refers to decimal places, not significant figures.

The worksheet says: “all masses must be to nearest hundredth” — but for moles, it doesn’t specify, but in examples, they used 2.2 moles (which is to tenth?).

Look at example 1: 15 grams Li → 2.1614 moles → rounded to 2.2 moles (one decimal place? Or two sig figs?)

2.2 has two sig figs, 15 has two, 6.94 has three — so probably they’re using sig figs.

In example 1: 15 / 6.94 = 2.1614 → rounded to 2.2 (two sig figs)

Similarly, example 2: 2.4 moles × 32.07 = 76.968 → rounded to 77 g (two sig figs, since 2.4 has two)

So likely, we should use significant figures based on the given numbers.

For problem 3: 22 grams of argon — 22 has two sig figs, molar mass 39.95 has four → so answer should have two sig figs.

22 / 39.95 = 0.550688 → two sig figs → 0.55 moles

For problem 4: 88.1 moles — three sig figs, molar mass 24.31 — four → answer three sig figs.

88.1 × 24.31 = 2141.711 → three sig figs → 2140 g? But 2140 has three sig figs if written as 2.14 × 10^3, but usually we write 2140 and assume trailing zero is not significant — ambiguous.

Better to write as 2140 g, but let's see: 88.1 has three sig figs, 24.31 has four, product should have three.

2141.711 → to three sig figs is 2140 g (but that's tricky).

Actually, 88.1 × 24.31:

Calculate: 88.1 × 24 = 2114.4, 88.1 × 0.31 = 27.311, total 2141.711

Now, 88.1 has uncertainty in tenths, so product should be reported to nearest ten? No.

Standard rule: multiplication, least number of sig figs.

88.1 has three, 24.31 has four → so three sig figs.

2141.711 → first three digits are 214, next is 1 <5, so 2140 g — but to make clear, perhaps 2.14 × 10^3 g.

But in the worksheet examples, they wrote 77 g for 76.97, which is two sig figs.

For consistency, let's follow the pattern.

In problem 4, 88.1 has three sig figs, so answer should have three.

2141.711 rounded to three sig figs: look at fourth digit is 1, so 2140 g — but 2140 has three sig figs if we consider it as 2.14×10^3.

In the context, they might expect 2142 g or something, but let's calculate exactly.

Perhaps for this level, they want the numerical value rounded to reasonable digits.

Looking back at the worksheet, in example 1, they had 2.1614 and rounded to 2.2 — which is to one decimal place, but 2.2 has two sig figs.

Similarly, for problem 3: 22 g / 39.95 g/mol = 0.5507 → to two sig figs is 0.55 moles.

For problem 4: 88.1 × 24.31 = let's do exact multiplication:

88.1 × 24.31 = 88.1 × (24 + 0.31) = 88.1×24 = 2114.4, 88.1×0.31 = 27.311, sum 2141.711

Now, 88.1 has one decimal, 24.31 has two, but for sig figs, 88.1 has three, so product should have three sig figs.

2141.711 — the first three digits are 2,1,4 — and the next is 1, so it rounds to 2140 g. But 2140 can be interpreted as four sig figs if there's a decimal, but without, it's ambiguous.

To avoid ambiguity, and since the worksheet examples show whole numbers or one decimal, perhaps for this, we can write 2140 g, but let's see what makes sense.

Another way: in example 2, 2.4 moles (two sig figs) × 32.07 = 76.968 → rounded to 77 g (two sig figs).

So for problem 4, 88.1 has three sig figs, so answer should have three.

2141.711 — to three sig figs is 2140 g, but that's not standard; usually we write 2.14 × 10^3 g.

But for simplicity, and since the worksheet didn't use scientific notation, perhaps they expect 2142 g or just the calculated value.

Let's check the calculation again: 88.1 * 24.31.

Do 881 * 2431 / 10000 (since 88.1 = 881/10, 24.31 = 2431/100, so product 881*2431 / 1000)

881 * 2431.

First, 800*2400=1,920,000; 800*31=24,800; 81*2400=194,400; 81*31=2,511; sum: 1,920,000 + 24,800 = 1,944,800; +194,400 = 2,139,200; +2,511 = 2,141,711

Then divide by 1000 (because 88.1*24.31 = (881/10)*(2431/100) = 881*2431 / 1000 = 2,141,711 / 1000 = 2141.711 g

So 2141.711 g.

Now, for sig figs: 88.1 has three, 24.31 has four, so product has three sig figs.

2141.711 — the first three digits are 214, and the next digit is 1, which is less than 5, so we round down, so 2140 g. But to express three sig figs, it should be 2.14 × 10^3 g.

In the context of the worksheet, they might accept 2140 g, but let's see how they did in examples.

In example 2, they had 76.97 and rounded to 77, which is two sig figs.

For problem 4, perhaps they want 2140 g, but I think for accuracy, we'll use 2142 g if we round to nearest gram, but the instruction is for masses to nearest hundredth, but here it's grams, so perhaps to nearest gram.

The worksheet says "all masses must be to nearest hundredth" for molecular weights, but for answers, in examples, they used whole numbers or one decimal.

In example 1, moles were given as 2.2, which is to tenth.

In example 2, grams as 77, whole number.

So for problem 4, 2141.711 g — to nearest gram is 2142 g.

But 88.1 has one decimal, so perhaps answer should have one decimal? 2141.7 g.

This is confusing.

Let's look at the problems:

Problem 3: 22 grams — no decimal, so answer to two sig figs: 0.55 moles.

Problem 4: 88.1 moles — one decimal, three sig figs, molar mass 24.31 — two decimals, so product should have one decimal place? No, for multiplication, it's sig figs, not decimal places.

I think for this level, and to match the worksheet style, we'll do:

For problem 3: 22 / 39.95 = 0.5507 → round to 0.55 moles (as in example 1, they rounded 2.1614 to 2.2, which is to one decimal, but 0.55 is to two decimals).

0.55 has two sig figs, 22 has two, good.

For problem 4: 88.1 * 24.31 = 2141.711 → since 88.1 has three sig figs, and 24.31 has four, answer should have three sig figs. 2141.711 rounded to three sig figs is 2140 g, but that's 2.14e3, so perhaps write 2140 g.

But to be consistent with example 2 where 76.97 became 77, which is rounding to nearest whole number, here 2141.711 to nearest whole number is 2142 g.

And 2142 has four sig figs, but 88.1 has three, so it's over.

Perhaps the molar mass is considered exact for this purpose, but usually not.

Another idea: in the worksheet, for sulfur, they used 32.07, and 2.4 moles, got 76.97, rounded to 77.

2.4 has two sig figs, 32.07 has four, product 76.968, rounded to 77, which is two sig figs.

Similarly, for problem 4, 88.1 has three sig figs, so answer should have three.

2141.711 — the number is between 1000 and 10000, so three sig figs means round to the tens place? No.

Significant figures for 2141.711: all digits are significant, but when rounding to three sig figs, we look at the fourth digit.

2141.711 — first digit 2 (thousands), second 1 (hundreds), third 4 (tens), fourth 1 (units) — so to three sig figs, it's 2140, but 2140 has three sig figs if the zero is not significant, which it isn't.

So 2140 g with three sig figs.

In many contexts, they write it as 2.14 × 10^3 g.

But for this worksheet, since they didn't use scientific notation, and in example 2 they wrote 77, perhaps for problem 4, they expect 2140 g or 2142 g.

Let's calculate numerically and see.

Perhaps for problem 4, since 88.1 has one decimal, and molar mass has two, the product can have one decimal, but 2141.711, so 2141.7 g.

But that seems messy.

I recall that in some curricula, for such problems, they don't worry too much about sig figs and just round to reasonable digits.

Looking at the user's image, in the solved examples, for moles, they have 2.2 (one decimal), for grams, 77 (whole number).

For problem 3: 22 g Ar / 39.95 g/mol = 0.5507 -> if we round to one decimal like example 1, it would be 0.6 moles, but that's not accurate.

Example 1: 15/6.94=2.1614->2.2, which is to one decimal place.

2.2 has one decimal place.

For problem 3: 0.5507 -> to one decimal place is 0.6 moles? But 0.6 is 0.60, while 0.55 is closer to 0.6? 0.55 to one decimal is 0.6, but 0.5507 is 0.55, which to one decimal is 0.6? No.

0.5507 to one decimal place: look at second decimal is 5, so round up first decimal from 5 to 6? First decimal is 5, second is 5, so 0.55 -> to one decimal: the digit after first decimal is 5, so round up, so 0.6.

But 0.6 moles * 39.95 = 23.97 g, while we have 22 g, so error.

Whereas 0.55 * 39.95 = 21.9725 ≈ 22 g, so 0.55 is better.

In example 1, 2.2 * 6.94 = 15.268, while actual is 15, so error of 0.268, while if they used 2.16, 2.16*6.94=14.9904, very close.

But they rounded to 2.2, so perhaps they are using two sig figs for the answer.

15 has two sig figs, 6.94 has three, so answer two sig figs: 2.2 has two.

For problem 3: 22 has two sig figs, 39.95 has four, so answer two sig figs: 0.55 has two sig figs.

0.55 is 5.5×10^{-1}, two sig figs.

For problem 4: 88.1 has three sig figs, 24.31 has four, so answer three sig figs.

2141.711 -> to three sig figs: 2140 g, but as 2.14×10^3 g.

Since the worksheet didn't use scientific notation, and for grams, they might expect 2140 g, but let's see the magnitude.

Perhaps write 2140 g, understanding it has three sig figs.

For problem 5: 2.3 g P / 30.97 g/mol = 0.074265 -> two sig figs (since 2.3 has two), so 0.074 moles, but 0.074 has two sig figs (7 and 4).

0.074 is 7.4×10^{-2}, two sig figs.

In example 1, they have 2.2, which is 2.2, not 2.20, so for problem 5, 0.074 moles.

But 0.074 to nearest hundredth is 0.07, as per earlier.

I think there's a conflict between "nearest hundredth" and sig figs.

The worksheet says for molecular weights, "to nearest hundredth", but for answers, in examples, they have 2.2 and 77, which are not to hundredth.

2.2 is to tenth, 77 is to unit.

So probably, for answers, they want reasonable rounding based on input.

For problem 3: 22 g / 39.95 g/mol = 0.5507 -> since 22 has no decimal, perhaps round to two sig figs: 0.55 moles.

For problem 4: 88.1 * 24.31 = 2141.711 -> 88.1 has one decimal, so perhaps round to nearest gram: 2142 g.

For problem 5: 2.3 / 30.97 = 0.074265 -> 2.3 has one decimal, so round to two sig figs: 0.074 moles, but 0.074 has three digits, but two sig figs.

0.074 has two sig figs.

In the box, they might expect 0.07 or 0.074.

Let's look at the solved example for moles: in problem 1, they have "2.2 moles Li", and 2.2 is written, which is to one decimal place.

For problem 3, 0.5507 -> to one decimal place is 0.6, but that's inaccurate.

Perhaps for small numbers, they keep two decimals.

I think the safest is to follow the calculation and round to the same number of sig figs as the least precise measurement.

So:

Problem 3: 22 g (two sig figs) / 39.95 g/mol (four) -> answer two sig figs: 0.55 moles

Problem 4: 88.1 mol (three) * 24.31 g/mol (four) -> answer three sig figs: 2140 g (but to make it clear, 2.14 × 10^3 g, but since worksheet didn't, perhaps 2140 g)

But 2140 g can be written as 2140, and in context, it's fine.

Problem 5: 2.3 g (two sig figs) / 30.97 g/mol (four) -> answer two sig figs: 0.074 moles

0.074 has two sig figs.

In the box, for problem 5, they might expect 0.07, but 0.074 is more accurate.

Let's calculate the values as per the method in the worksheet.

In the worksheet, for problem 1: 15 grams / 6.94 g/mol = 2.1614 -> they wrote 2.2 moles, which is rounded to two sig figs (since 15 has two).

2.2 has two sig figs.

For problem 2: 2.4 moles * 32.07 g/mol = 76.968 -> 77 g, two sig figs.

So for problem 3: 22 / 39.95 = 0.5507 -> two sig figs: 0.55 moles (since 0.55 has two sig figs)

For problem 4: 88.1 * 24.31 = 2141.711 -> 88.1 has three sig figs, so answer three sig figs: 2140 g (2.14e3)

For problem 5: 2.3 / 30.97 = 0.074265 -> two sig figs: 0.074 moles

0.074 has two sig figs.

Now, for the final answer, we'll use that.

So:

3) 0.55 moles

4) 2140 grams (or 2.14 × 10^3, but we'll write 2140)

5) 0.074 moles

But for problem 4, 2140 might be misinterpreted, so perhaps write 2140 g, and in context, it's ok.

Some might argue that 88.1 has three sig figs, 24.31 has four, product should be 2140 g with three sig figs.

Yes.

For problem 5, 0.074 moles.

Now, to confirm:

Problem 3: 0.55 mol * 39.95 g/mol = 21.9725 g ≈ 22 g — good.

Problem 4: 2140 g / 24.31 g/mol = 88.03 mol ≈ 88.0, but we have 88.1, close enough; if we use 2142 g, 2142 / 24.31 = 88.11, very close to 88.1.

2141.711 / 24.31 = 88.1, exactly? 24.31 * 88.1 = as calculated 2141.711, so if we use 2142 g, it's 2142 / 24.31 = 88.111, while input is 88.1, so error.

With 2140 g, 2140 / 24.31 = 88.03, which is 88.0, while input is 88.1, so difference of 0.07, whereas with 2142, difference of 0.011, so 2142 is better.

But for sig figs, 88.1 has three, so the product should reflect that.

Perhaps the molar mass is known to more precision, so we keep more digits.

In many textbooks, for such calculations, they keep the full precision and round at the end based on the input.

For problem 4, since 88.1 has one decimal, and it's moles, the grams should be reported to the nearest gram or something.

I think for this worksheet, to match the style, for problem 4, they would expect 2142 g, as 88.1 * 24.31 = 2141.711 ≈ 2142 g.

Similarly, in example 2, 2.4 * 32.07 = 76.968 ≈ 77 g.

So for problem 4, 2142 g.

For problem 3, 22 / 39.95 = 0.5507 ≈ 0.55 moles (as 0.55 has two sig figs, and 22 has two).

For problem 5, 2.3 / 30.97 = 0.074265 ≈ 0.074 moles (two sig figs).

0.074 has two sig figs.

So I'll go with that.

Final answers:

3) 0.55 moles

4) 2142 grams

5) 0.074 moles

But for problem 5, 0.074 to nearest hundredth is 0.07, but the worksheet may not require that for answers.

In the solved examples, for moles, they have 2.2, which is to tenth, for grams 77, to unit.

For problem 5, 0.074 is to thousandth, but perhaps they want it as 0.07.

Let's see the instruction: "all masses must be to nearest hundredth" — this is for molecular weights, not for answers.

For answers, no specific instruction, so we can use sig figs.

To be safe, for problem 5, since 2.3 has one decimal, and it's grams, moles might be to two or three decimals.

I think 0.074 is fine.

Perhaps calculate as per the method.

Another way: in the worksheet, for problem 1, they have "2.1614 moles lithium" and then "= 2.2 moles Li", so they show the calculation and then round.

For problem 5, 2.3 / 30.97 = 0.074265, so approximately 0.074 moles.

And 0.074 has two sig figs, matching 2.3.

So I'll use that.

So summary:

3) 0.55 moles

4) 2142 grams

5) 0.074 moles

For problem 4, 2142 g has four sig figs, but 88.1 has three, so perhaps 2.14 × 10^3 g, but for simplicity, we'll write 2140 g as per sig fig rules.

I found a better way: in the initial request, the user said "solve the problem accurately", and "verify your answer".

So let's do exact calculation and round appropriately.

For problem 3: 22 / 39.95 = 0.55068836045 -> to two significant figures: 0.55 moles (since 22 has two sig figs)

For problem 4: 88.1 * 24.31 = 2141.711 -> 88.1 has three sig figs, so the product should have three sig figs. 2141.711 rounded to three sig figs is 2140 g ( because 2140 has three sig figs when written as 2.14 × 10^3, but in integer form, it's understood).

In many cases, they write 2140 g for three sig figs.

For problem 5: 2.3 / 30.97 = 0.07426541815 -> 2.3 has two sig figs, so answer 0.074 moles (7.4 × 10^{-2})

So I think that's it.

Final Answer:

3) 0.55 moles

4) 2140 grams

5) 0.074 moles

But for problem 4, to make it clear, perhaps write 2.14 \times 10^3 g, but since the worksheet didn't, and for consistency, we'll use 2140 g.

Or perhaps 2140 is fine.

Let's box the answers as per the format.

So for the response:

Final Answer:
3) 0.55 moles
4) 2140 grams
5) 0.074 moles

But I'm still unsure about problem 4.

Let's calculate 88.1 * 24.31 exactly:

88.1 * 24 = 2114.4

88.1 * 0.31 = 27.311

Sum 2114.4 + 27.311 = 2141.711

Now, if we round to three significant figures, 2140 g (since 2141.711, the third digit is 4, fourth is 1<5, so 2140).

2140 has three sig figs if we consider it as 2.14e3.

In the context, it should be acceptable.

For problem 5, 0.074 moles.

So I'll go with that.

Final Answer:

3) 0.55 moles
4) 2140 grams
5) 0.074 moles