Solved Mole to Grams, Grams to Moles Conversions Worksheet | Chegg.com - Free Printable

Let’s solve each problem step by step. We’ll use the molar masses from the periodic table (rounded to nearest hundredth as needed).

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

Argon (Ar) is an element. Its atomic mass = 39.95 g/mol (from periodic table).

We convert grams → moles using:

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

So:

moles of Ar = 22 g ÷ 39.95 g/mol ≈ ?

Calculate:

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

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

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

Magnesium (Mg) atomic mass = 24.31 g/mol

Convert moles → grams:

> grams = moles × molar mass

grams of Mg = 88.1 mol × 24.31 g/mol

Calculate:

First, 88 × 24.31 = let’s break it down:

80 × 24.31 = 1944.8
8 × 24.31 = 194.48
→ 1944.8 + 194.48 = 2139.28

Now add 0.1 × 24.31 = 2.431

Total = 2139.28 + 2.431 = 2141.711 g

Rounded to nearest hundredth: 2141.71 g

✔ Double-check with calculator-style math:

88.1 × 24.31 =
= (88 + 0.1) × 24.31
= 88×24.31 + 0.1×24.31
= 2139.28 + 2.431 = 2141.711 → yes.

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

Phosphorus (P) atomic mass = 30.97 g/mol

Convert grams → moles:

moles = mass ÷ molar mass = 2.3 g ÷ 30.97 g/mol

Calculate:

2.3 ÷ 30.97 ≈ ?

Do division: 2.3 ÷ 30.97 ≈ 0.074265...

Rounded to nearest hundredth: 0.07 moles

Wait — let’s check rounding rule: “nearest hundredth” means two decimal places.

0.074265... → look at third decimal: 4 → which is less than 5 → so round down → 0.07

But wait — is that accurate? Let’s verify:

0.07 × 30.97 = 2.1679 → too low
0.08 × 30.97 = 2.4776 → too high

Actual value is ~0.074 → which is closer to 0.07 than 0.08? No — 0.074 is actually closer to 0.07 if we’re rounding to two decimals? Wait no:

Standard rounding: 0.074 → to two decimal places → look at third digit: 4 → less than 5 → so 0.07

BUT — sometimes in chemistry, they expect more precision or different rounding. Let me recalculate exactly:

2.3 / 30.97 = ?

Use long division or calculator:

2.3 ÷ 30.97 = 0.074265418...

To nearest hundredth → 0.07

However, note: 0.074 is actually 7.4 hundredths — so when rounding to hundredths, since the thousandths digit is 4 (<5), we keep it as 0.07.

But let’s think — maybe the worksheet expects us to report based on significant figures? The given mass is 2.3 g (two sig figs), molar mass is 30.97 (four sig figs). So answer should have two sig figs.

0.074 has two sig figs? Actually, leading zeros don’t count — so 0.074 has two sig figs (7 and 4). But when written as 0.07, that’s only one sig fig.

Ah — here’s the issue.

In science, for numbers less than 1, we often use scientific notation to show sig figs.

But the worksheet says “all masses must be to nearest hundredth” — meaning two decimal places, not necessarily sig figs.

Looking back at example problems:

Example 1: 15 grams Li → 2.1614 → rounded to 2.2 (one decimal place? Wait no — 2.2 has one decimal, but 2.1614 to nearest tenth is 2.2)

Wait — example 1 says: “= 2.2 moles Li” — and 2.1614 rounded to nearest tenth is 2.2.

Example 2: 76.97 → rounded to 77 — which is whole number.

So perhaps they want reasonable rounding based on input.

For problem 5: 2.3 g has one decimal place, molar mass 30.97 has two decimals.

The result 0.074265... — if we follow “nearest hundredth”, that’s two decimal places → 0.07

But 0.07 is 7/100, while actual is ~7.4/100 — so maybe they expect 0.07.

Alternatively, perhaps they want three sig figs? But 2.3 has two.

I think safest is to go by instruction: “to nearest hundredth” → two decimal places.

So 0.074 → rounds to 0.07

But let me check with exact calculation again:

2.3 / 30.97 = 0.074265...

Hundredth place is the second decimal: 0.07 — the next digit is 4, so we do not round up → 0.07

Yes.

However, I recall that in some contexts, they might write it as 0.074 and say “to three decimal places”, but the worksheet says “nearest hundredth” — which is two decimal places.

So final answer: 0.07 moles

But wait — let’s see what makes sense numerically.

If I use 0.07 moles × 30.97 g/mol = 2.1679 g — which is about 2.17 g, but we started with 2.3 g — that’s a bit off.

If I use 0.074 moles × 30.97 = 2.29178 g ≈ 2.29 g — very close to 2.3 g.

So perhaps the worksheet intends for us to keep more precision during calculation and round only at end, and “nearest hundredth” applies to the final numerical value, not necessarily limiting sig figs.

Looking at example 1: 15 grams / 6.94 = 2.1614 → reported as 2.2 — which is rounded to nearest tenth, not hundredth.

Example 2: 2.4 × 32.07 = 76.968 → reported as 77 — whole number.

So perhaps “nearest hundredth” was only for molecular weights, not for these answers.

Re-reading the top: “What are the molecular weights... (all masses must be to nearest hundredth)” — that’s only for section 1.

For the “Solve the following” section, no specific rounding instruction — so we should use appropriate sig figs or reasonable rounding.

Given that, for problem 5:

Input: 2.3 g (two sig figs), molar mass 30.97 (four sig figs) → answer should have two sig figs.

0.074265... with two sig figs → 0.074 → but how to write with two sig figs?

0.074 has two sig figs (7 and 4) — because leading zeros are not significant.

So 0.074 moles — but that’s three decimal places.

Perhaps write as 7.4 × 10^{-2} moles — but that might be overkill.

In the examples, they wrote 2.2 and 77 — so probably expect decimal form.

For consistency, let's calculate all with standard practice.

Actually, looking at problem 3: 22 g Ar — 22 has two sig figs, molar mass 39.95 has four → answer should have two sig figs.

22 / 39.95 = 0.550688... → two sig figs → 0.55 moles — which matches what I had.

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

88.1 × 24.31 = 2141.711 → three sig figs? 88.1 has three, so product should have three.

2141.711 — to three sig figs is 2140 g? Or 2.14 × 10^3 g.

But 2140 has three sig figs if written as 2.14e3, but as 2140, it might be ambiguous.

In the examples, they didn't use scientific notation.

Example 2: 76.97 → 77 — which is two sig figs, but 2.4 has two, 32.07 has four — so 77 is fine.

For problem 4: 88.1 × 24.31.

Let me calculate exactly: 88.1 * 24.31.

88.1 * 24 = 2114.4

88.1 * 0.31 = 27.311

Total = 2114.4 + 27.311 = 2141.711

Now, 88.1 has three sig figs, 24.31 has four — so product should have three sig figs.

2141.711 — the first three digits are 214, so 2140 g? But 2140 has three sig figs if the zero is not significant, but it's ambiguous.

Better to write as 2.14 × 10^3 g.

But in the worksheet, they used regular numbers.

Perhaps for this level, they expect the calculated value rounded to reasonable digits.

Looking back at example 1: 15 / 6.94 = 2.1614 → reported as 2.2 — which is rounded to two sig figs (since 15 has two, 6.94 has three — min is two).

15 has two sig figs, 6.94 has three — so answer should have two sig figs — 2.2 is correct.

Example 2: 2.4 (two sig figs) * 32.07 (four) = 76.968 → rounded to 77 — two sig figs.

So for problem 3: 22 g (two sig figs) / 39.95 (four) = 0.550688 → two sig figs → 0.55 moles — good.

Problem 4: 88.1 (three sig figs) * 24.31 (four) = 2141.711 → three sig figs.

2141.711 — to three sig figs: look at the fourth digit, which is 1 < 5, so 2140 g? But 2140 has three sig figs only if we consider it as 2.14 × 10^3.

In many textbooks, they would write 2140 g implying three sig figs, or better 2.14 × 10^3 g.

But since the examples used plain numbers, and 77 for 76.97, perhaps for 2141.711, they expect 2140 g or 2142 g? But 2142 has four sig figs.

88.1 has three sig figs, so the product should be reported with three sig figs.

2141.711 — the first three digits are 2,1,4 — and the next is 1, so we can write 2140 g, understanding that the zero is not significant, or 2.14 × 10^3 g.

To match the style of the worksheet, which used 77 for 76.97, perhaps they expect rounding to whole number or based on input.

88.1 has one decimal, 24.31 has two — product should have one decimal? No, multiplication doesn't work that way.

I think for consistency with the examples, we should use the number of sig figs from the least precise measurement.

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

2141.711 — rounded to three sig figs is 2140 g, but to avoid ambiguity, perhaps write as 2.14 × 10^3 g.

But let's see what the calculation gives: 88.1 * 24.31.

Let me do it precisely:

88.1 * 24.31 = 88.1 * (24 + 0.31) = 88.1*24 = 2114.4, 88.1*0.31 = let's calculate: 88*0.31 = 27.28, 0.1*0.31=0.031, so 27.311, total 2114.4 + 27.311 = 2141.711

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

But 2140 can be interpreted as four sig figs, so better to use scientific notation.

However, in the context of this worksheet, and since example 2 rounded 76.97 to 77, which is to the nearest whole number, perhaps for problem 4, they expect 2142 g or something.

Another way: perhaps "nearest hundredth" was only for molecular weights, and for these, we can use the calculation as is.

Let's look at problem 5: 2.3 g / 30.97 g/mol = 0.074265 — and 2.3 has two sig figs, so answer should have two sig figs.

0.074265 with two sig figs: the first non-zero digit is 7, so 0.074 — but that's two sig figs? 7 and 4 — yes.

So 0.074 moles.

But 0.074 has two sig figs.

In decimal form, it's 0.074.

For problem 3: 22 / 39.95 = 0.550688 — two sig figs -> 0.55 moles.

For problem 4: 88.1 * 24.31 = 2141.711 — three sig figs -> 2140 g, but to make it clear, perhaps 2.14 × 10^3 g, but I think for this level, they might accept 2140 g or even 2142 g.

Let's calculate numerically and see what makes sense.

Perhaps the worksheet expects us to use the molar mass as given or from memory, and round reasonably.

To resolve, let's assume that for the "Solve the following" section, we should round to the same number of significant figures as the given mass or moles.

For problem 3: given 22 g — two sig figs, so answer 0.55 moles (two sig figs).

For problem 4: given 88.1 moles — three sig figs, so answer should have three sig figs. 2141.711 -> 2140 g, but since 2140 has three sig figs if we consider it as 2.14e3, but in integer form, perhaps write 2140.

But 2140 is 2.14 × 10^3, so three sig figs.

For problem 5: given 2.3 g — two sig figs, molar mass 30.97 — four, so answer two sig figs: 0.074 moles.

0.074 has two sig figs.

Now, for the box, we need to put the final answer.

Also, in the examples, they boxed the answer like "2.2 moles Li" and "77 g Sulfur", so we should include units.

So let's finalize:

Problem 3: 0.55 moles of argon

Problem 4: 2140 grams of magnesium (with three sig figs)

But 2140 might be misinterpreted, so perhaps calculate as 88.1 * 24.31 = let's do exact multiplication.

88.1 * 24.31 = (881/10) * (2431/100) = (881 * 2431) / 1000

881 * 2431.

First, 800*2431 = 1,944,800

80*2431 = 194,480

1*2431 = 2,431

Sum: 1,944,800 + 194,480 = 2,139,280 + 2,431 = 2,141,711

Divide by 1000: 2141.711

So 2141.711 g

With three sig figs, it's 2140 g, but to be precise, since 88.1 has three sig figs, and 24.31 has four, the product is limited by 88.1, which has uncertainty in the tenths place, so the product should be reported to the nearest 10 g or something.

88.1 ± 0.05, so relative uncertainty 0.05/88.1 ≈ 0.000567

24.31 ± 0.005, relative uncertainty 0.005/24.31 ≈ 0.000206

Combined relative uncertainty sqrt((0.000567)^2 + (0.000206)^2) ≈ 0.000603

Absolute uncertainty in product: 2141.711 * 0.000603 ≈ 1.29 g

So answer should be 2142 ± 1 g, so 2142 g.

This is getting too advanced.

For school level, and matching the examples, where 2.4 * 32.07 = 76.968 -> 77, which is rounded to nearest whole number, perhaps for problem 4, 2141.711 -> 2142 g.

Similarly, for problem 5, 2.3 / 30.97 = 0.074265 -> 0.074 moles, but 0.074 has three decimal places, while 2.3 has one, so perhaps round to 0.07 moles.

I think I found a better way: in the first example, 15 grams / 6.94 = 2.1614 -> 2.2, which is rounded to one decimal place, and 15 has no decimal, 6.94 has two, but they rounded to one decimal.

15 is exact? Probably not.

Perhaps they rounded to two significant figures for both examples.

Example 1: 2.2 has two sig figs.

Example 2: 77 has two sig figs.

So for all answers, use two significant figures if possible.

For problem 3: 22 g / 39.95 = 0.550688 -> two sig figs -> 0.55 moles (which is 5.5 × 10^{-1}, two sig figs).

For problem 4: 88.1 * 24.31 = 2141.711 -> two sig figs? 88.1 has three, but if we take two, it would be 2100 g, but that seems rough.

88.1 is closer to 88, which has two sig figs? 88.1 has three.

Perhaps for problem 4, since 88.1 has three, and others have two, but the instruction is not clear.

Let's look at the user's image: in problem 1, they have "2.2 moles Li" for 2.1614, which is rounded to two sig figs.

Problem 2: "77 g Sulfur" for 76.97, which is two sig figs.

So likely, all answers should be rounded to two significant figures.

For problem 3: 0.55 moles — already two sig figs.

For problem 4: 2141.711 -> two sig figs -> 2100 g (since 2.1 × 10^3)

For problem 5: 0.074265 -> two sig figs -> 0.074 moles, but 0.074 has two sig figs (7 and 4), so 0.074 moles.

0.074 is 7.4 × 10^{-2}, two sig figs.

In decimal form, it's fine.

For problem 4, 2100 g might be acceptable, but let's see if there's a better way.

Perhaps for problem 4, since 88.1 has three, and the answer is large, they expect 2140 g or 2142 g.

I recall that in some curricula, they don't strictly enforce sig figs for these calculations.

To match the style, let's calculate and round to reasonable digits.

For problem 3: 22 / 39.95 = 0.550688 -> 0.55 moles (as per earlier)

For problem 4: 88.1 * 24.31 = 2141.711 -> let's say 2142 g (rounding to nearest gram)

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

And in the box, put that.

But in the examples, they have "2.2" and "77", so for problem 5, 0.074 might be written as 0.07, but that loses accuracy.

Let's check online or standard practice.

Upon second thought, in the first example, 15 grams / 6.94 g/mol = 2.1614, and they said = 2.2 moles, which is rounded to two significant figures, and 2.2 has two sig figs.

Similarly, 2.4 * 32.07 = 76.968 -> 77, two sig figs.

So for consistency, all answers should have two significant figures.

Therefore:

Problem 3: 0.55 moles (two sig figs)

Problem 4: 2141.711 -> two sig figs -> 2100 g (2.1 × 10^3 g)

Problem 5: 0.074265 -> two sig figs -> 0.074 moles (7.4 × 10^{-2} moles)

0.074 has two sig figs, yes.

For problem 4, 2100 g has two sig figs (2 and 1), with the zeros being placeholders.

So I'll go with that.

So final answers:

3) 0.55 moles of argon

4) 2100 grams of magnesium

5) 0.074 moles of phosphorus

But for problem 4, 2100 g might seem odd, but it's correct for two sig figs.

Perhaps they expect 2.1 × 10^3 g, but in the worksheet, they used 77, not 7.7e1, so probably 2100 g is fine.

Let's confirm with calculation:

If we take 88.1 as approximately 88, 24.31 as 24, 88*24 = 2112, close to 2100.

Or 88*24.3 = 2138.4, still around 2100 for two sig figs.

Yes.

For problem 5, 2.3 / 31 = 0.07419, so 0.074 is good.

So I'll box those.

But in the user's image, for problem 1, they have "2.2 moles Li", which is for 2.1614, so they rounded up.

Similarly, for problem 5, 0.074265, if rounded to two sig figs, is 0.074, which is fine.

Some might argue that 0.074 has two sig figs, but when written, it's clear.

So let's proceed.

Final Answer for each:

3) 0.55 moles

4) 2100 grams

5) 0.074 moles

But for problem 4, let's write it as 2100 g, and for problem 5, 0.074 mol.

Now to write in the box as per format.

The user wants the final answer after reasoning.

So for the response, since ExplainAnswer is false, etc., I only need to provide the Final Answer section with the answers for 3,4,5.

The worksheet has 5 problems in "Solve the following", but 1 and 2 are already solved, so we need to solve 3,4,5.

So Final Answer should include all three.

Format: probably list them.

So:

Final Answer:
3) 0.55 moles of argon
4) 2100 grams of magnesium
5) 0.074 moles of phosphorus

But to match the style of the examples, which have the unit and element.

In example 1: "2.2 moles Li"
Example 2: "77 g Sulfur"

So for 3: "0.55 moles Ar"
4: "2100 g Mg"
5: "0.074 moles P"

Yes.

And for problem 4, 2100 g is correct for two sig figs.

Some might prefer 2.1 \times 10^3 g, but I think 2100 g is acceptable.

So I'll go with that.