Empirical and Molecular Formula Worksheet with handwritten solutions and chemical notations.

Let’s solve each problem step by step. We’re finding empirical formulas — the simplest whole-number ratio of atoms in a compound.

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Problem 1:

We are given masses or moles and need to find the empirical formula.

#### Part a:
Given:
- C = 0.783 g
- H = 0.196 g
- O = 0.521 g

Step 1: Convert grams to moles using atomic masses:
- C: 0.783 g ÷ 12.01 g/mol ≈ 0.0652 mol
- H: 0.196 g ÷ 1.008 g/mol ≈ 0.1944 mol
- O: 0.521 g ÷ 16.00 g/mol ≈ 0.0326 mol

Step 2: Divide each by the smallest number of moles (which is 0.0326):
- C: 0.0652 ÷ 0.0326 ≈ 2.00
- H: 0.1944 ÷ 0.0326 ≈ 5.96 → ~6
- O: 0.0326 ÷ 0.0326 = 1

So, ratio is C₂H₆O → Empirical formula: C₂H₆O

*(Note: The handwritten answer says C₄H₁₀O₂ — that’s actually the molecular formula if molar mass is double. But since we’re asked for empirical, it should be simplified to C₂H₆O. However, looking at the work shown, they may have used different rounding or intended molecular. Let’s check their math.)*

Wait — let’s recheck with exact values:

C: 0.783 / 12.01 = 0.065195...
H: 0.196 / 1.008 = 0.194444...
O: 0.521 / 16.00 = 0.0325625

Divide by 0.0325625:

C: 0.065195 / 0.0325625 ≈ 2.002 → 2
H: 0.194444 / 0.0325625 ≈ 5.97 → 6
O: 1

Still C₂H₆O.

But in the image, they wrote C₄H₁₀O₂ — which suggests they might have multiplied by 2? That would be incorrect for empirical unless specified. Actually, wait — maybe they made an error in division?

Looking again: In the image, under part a, they have:

They calculated moles as:
C: 0.783/12 = 0.06525 → then divided by 0.0326 → got 2.00
H: 0.196/1 = 0.196 → /0.0326 = 6.01 → rounded to 6
O: 0.521/16 = 0.03256 → /0.03256 = 1

Then they wrote “C₂H₆O” but then crossed out and wrote “C₄H₁₀O₂”? Wait no — actually in the box they wrote C₄H₁₀O₂. That must be a mistake.

Actually, let me recalculate based on what they did:

In the image, they have:

For C: 0.783 / 12 = 0.06525 → then they say “divide by 0.0326” → 2.00
H: 0.196 / 1 = 0.196 → /0.0326 = 6.01 → 6
O: 0.521 / 16 = 0.03256 → /0.03256 = 1

So ratio is 2:6:1 → C₂H₆O

But then they wrote C₄H₁₀O₂ — which is not correct for empirical. Unless... perhaps they meant molecular? But the question says "empirical".

Wait — looking back at the original problem statement in the image: It says “Determine the empirical formula...”

So C₂H₆O is correct.

But in the boxed answer, they have C₄H₁₀O₂ — that’s wrong for empirical. Maybe typo?

Actually, let’s check part b — maybe I misread.

No — for part a, the correct empirical formula is C₂H₆O

But let’s move on and come back.

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#### Part b:
Given:
- Ag = 63.5%
- S = 8.2%
- O = 28.3%

Assume 100g sample → so masses are:
Ag: 63.5 g
S: 8.2 g
O: 28.3 g

Convert to moles:
Ag: 63.5 / 107.87 ≈ 0.5887 mol
S: 8.2 / 32.07 ≈ 0.2557 mol
O: 28.3 / 16.00 ≈ 1.7688 mol

Divide by smallest (0.2557):
Ag: 0.5887 / 0.2557 ≈ 2.30
S: 0.2557 / 0.2557 = 1
O: 1.7688 / 0.2557 ≈ 6.92

Not whole numbers. Multiply by 3 to get integers:
Ag: 2.30 × 3 = 6.9 → ~7
S: 1 × 3 = 3
O: 6.92 × 3 = 20.76 → ~21

That gives Ag₇S₃O₂₁ — which simplifies? Not really. But that seems messy.

Wait — perhaps calculation error.

Let me use more precise atomic masses:

Ag: 107.87 g/mol
S: 32.065 g/mol
O: 16.00 g/mol

Moles:
Ag: 63.5 / 107.87 = 0.5887
S: 8.2 / 32.065 = 0.2557
O: 28.3 / 16.00 = 1.76875

Now divide by S moles (0.2557):

Ag: 0.5887 / 0.2557 = 2.302
O: 1.76875 / 0.2557 = 6.917

So ratios: Ag : S : O = 2.302 : 1 : 6.917

Multiply by 3:
Ag: 6.906 → 7
S: 3
O: 20.75 → 21

So Ag₇S₃O₂₁ — but this can be simplified? GCD of 7,3,21 is 1 — so empirical formula is Ag₇S₃O₂₁? That seems unusual.

But in the image, they have AgNO₃ — which is silver nitrate, but that has N, not S. Wait — the problem says S (sulfur), but they wrote AgNO₃? That must be a mistake.

Looking at the image: For part b, they have:

They wrote:
Ag: 63.5 / 107.9 = 0.588
S: 8.2 / 32.1 = 0.255
O: 28.3 / 16 = 1.769

Then divided by 0.255:
Ag: 2.306
S: 1
O: 6.937

Then they said multiply by 3:
Ag: 6.92 → 7
S: 3
O: 20.8 → 21

But then they wrote AgNO₃ — which is wrong because there's no nitrogen! The element is sulfur (S), not nitrogen (N).

This is a critical error. The compound contains Ag, S, O — so it should be something like Ag₂SO₄ or similar.

Let me try assuming it's Ag₂SO₄:

Molar mass: 2*107.87 + 32.06 + 4*16 = 215.74 + 32.06 + 64 = 311.8 g/mol

Percentages:
Ag: 215.74/311.8 ≈ 69.2% — but given is 63.5% — too high.

Try Ag₂S₂O₃? Or perhaps Ag₂SO₃?

Ag₂SO₃: 2*107.87 + 32.06 + 48 = 215.74 + 32.06 + 48 = 295.8

Ag%: 215.74/295.8 ≈ 72.9% — still too high.

Perhaps it's Ag₂S? But no oxygen.

Another idea: maybe the percentages are for Ag, N, O? But the problem says S.

Looking back at the user's image: In part b, it says "Ag, S, and O" — but in their calculation, they have "AgNO3" — which suggests they misread S as N.

That must be it. Probably a typo in the problem or in their reading.

But according to the text, it's S (sulfur). So we must proceed with S.

From earlier: ratios Ag:S:O = 2.302:1:6.917

Let me try multiplying by 2:
Ag: 4.604
S: 2
O: 13.834 — not good.

Multiply by 3: Ag:6.906, S:3, O:20.751 — close to 7:3:21

7:3:21 can be divided by... gcd is 1, so empirical formula is Ag₇S₃O₂₁

But that's unusual. Perhaps it's (Ag₂SO₄)₃.5 or something — not integer.

Maybe calculate exactly:

Let me set up equations.

Let the formula be Ag_x S_y O_z

Then:
( x * 107.87 ) / total mass = 0.635
( y * 32.065 ) / total mass = 0.082
( z * 16.00 ) / total mass = 0.283

From second equation: y * 32.065 = 0.082 * total mass
From first: x * 107.87 = 0.635 * total mass

Divide first by second:
(x * 107.87) / (y * 32.065) = 0.635 / 0.082 = 7.7439

So x/y = (7.7439 * 32.065) / 107.87 = (248.3) / 107.87 ≈ 2.302

Same as before.

Similarly, from third and second:
(z * 16) / (y * 32.065) = 0.283 / 0.082 = 3.4512

So z/y = (3.4512 * 32.065) / 16 = (110.66) / 16 = 6.916

So x:y:z = 2.302 : 1 : 6.916

To get integers, multiply by 3: 6.906 : 3 : 20.748 — very close to 7:3:21

So empirical formula is Ag₇S₃O₂₁

But this can be written as Ag₇(SO₄)₃· something? SO₄ is S O4, so S₃O₁₂, but we have O₂₁, so extra O9 — not matching.

Perhaps it's Ag₂S₂O₇ or other.

Another common compound is silver sulfate Ag₂SO₄, but as above, percentage doesn't match.

Let me calculate what percentage Ag₂SO₄ would have:

Molar mass Ag₂SO₄ = 2*107.87 + 32.06 + 64 = 215.74 + 32.06 + 64 = 311.8 g/mol

Ag% = 215.74/311.8 * 100 = 69.19% — but given is 63.5%, so not match.

Try Ag₂S₂O₃ (silver thiosulfate?): 2*107.87 + 2*32.06 + 3*16 = 215.74 + 64.12 + 48 = 327.86

Ag% = 215.74/327.86 * 100 = 65.8% — closer to 63.5%? Still higher.

65.8 vs 63.5 — not great.

Try Ag₂S: 2*107.87 + 32.06 = 247.8, Ag% = 215.74/247.8 = 87.06% — too high.

Perhaps it's not silver sulfite or sulfate.

Another idea: maybe the "S" is a typo and it's "N" for nitrogen. Because in the image, they have AgNO3, which is common.

Let me check if it makes sense with N.

If it's Ag, N, O:

Given: Ag 63.5%, N 8.2%, O 28.3% — but nitrogen atomic mass is 14, not 32.

Moles:
Ag: 63.5 / 107.87 = 0.5887
N: 8.2 / 14.01 = 0.5853
O: 28.3 / 16 = 1.76875

Divide by smallest (0.5853):
Ag: 0.5887 / 0.5853 ≈ 1.006
N: 0.5853 / 0.5853 = 1
O: 1.76875 / 0.5853 ≈ 3.022

So approximately 1:1:3 → AgNO₃

And percentages for AgNO₃:
Molar mass = 107.87 + 14.01 + 48 = 169.88
Ag% = 107.87/169.88 * 100 = 63.5%
N% = 14.01/169.88 * 100 = 8.24%
O% = 48/169.88 * 100 = 28.25% — matches perfectly!

So the problem likely has a typo: it should be "N" (nitrogen), not "S" (sulfur). Because with S, it doesn't give a nice formula, but with N, it gives AgNO₃ exactly.

In the image, they also have AgNO₃ boxed, so probably it's intended to be nitrogen.

I think we should go with AgNO₃ for part b, assuming "S" was a typo and should be "N".

Otherwise, if strictly S, it's Ag₇S₃O₂₁, but that's unlikely for a homework problem.

So I'll assume it's N.

Empirical formula: AgNO₃

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#### Part c:
Given:
- A 1.078 g sample of a gas found to contain 0.540 g of sulfur and 0.538 g of oxygen.

Masses: S = 0.540 g, O = 0.538 g

Moles:
S: 0.540 / 32.065 ≈ 0.01684 mol
O: 0.538 / 16.00 = 0.033625 mol

Divide by smallest (0.01684):
S: 0.01684 / 0.01684 = 1
O: 0.033625 / 0.01684 ≈ 1.997 → 2

So ratio 1:2 → empirical formula SO₂

In the image, they have SO₂ boxed — correct.

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#### Part d:
Given:
- A 0.5998 g sample of a new compound was analyzed and found to contain 0.2322 g of carbon, 0.05848 g of hydrogen, and 0.3091 g of oxygen.

Masses: C=0.2322g, H=0.05848g, O=0.3091g

Moles:
C: 0.2322 / 12.01 ≈ 0.01933 mol
H: 0.05848 / 1.008 ≈ 0.05802 mol
O: 0.3091 / 16.00 = 0.01932 mol

Divide by smallest (0.01932):
C: 0.01933 / 0.01932 ≈ 1.0005 → 1
H: 0.05802 / 0.01932 ≈ 3.003 → 3
O: 0.01932 / 0.01932 = 1

So ratio 1:3:1 → empirical formula CH₃O

But CH₃O is not standard; usually written as CH₃OH for methanol, but empirical is CH₄O? Wait.

CH₃O would imply C H3 O, but valence: carbon has 4 bonds, in CH₃O, if it's radical, but typically empirical formula for methanol is CH₄O.

Let me check the moles again.

C: 0.2322 / 12.01 = let's calculate precisely: 0.2322 ÷ 12.01 = 0.019333888...
H: 0.05848 ÷ 1.008 = 0.058015873...
O: 0.3091 ÷ 16.00 = 0.01931875

Smallest is O: 0.01931875

C: 0.019333888 / 0.01931875 ≈ 1.00078 → 1
H: 0.058015873 / 0.01931875 ≈ 3.003 → 3
O: 1

So C₁H₃O₁ — but this is not a stable compound; usually it's CH₄O for methanol.

Perhaps rounding error.

Total mass: 0.2322 + 0.05848 + 0.3091 = 0.59978 g — matches sample mass.

Now, if empirical formula is CH₃O, molar mass would be 12+3+16=31 g/mol.

But for methanol, it's CH₄O, 32 g/mol.

Check the hydrogen mass: for CH₄O, H should be 4/32 = 12.5% of mass.

Here, H mass = 0.05848 / 0.5998 ≈ 9.75% — while for CH₄O, it should be 4*1.008/32.04 ≈ 4.032/32.04 ≈ 12.58% — not match.

For CH₃O: 3*1.008 / 31.024 ≈ 3.024/31.024 ≈ 9.75% — yes! Matches.

0.05848 / 0.5998 = ? Calculate: 0.05848 ÷ 0.5998 ≈ 0.0975 — 9.75%

For CH₃O: H% = 3.024 / (12.01 + 3.024 + 16.00) = 3.024 / 31.034 ≈ 0.09745 — 9.745% — matches.

C%: 12.01/31.034 ≈ 38.7%, here 0.2322/0.5998≈38.71% — good.

O%: 16/31.034≈51.56%, here 0.3091/0.5998≈51.53% — good.

So empirical formula is CH₃O

But typically, we write it as CH₃O, though it's often associated with methoxy group, but for empirical, it's fine.

In some contexts, it might be written as CH₄O if it's methanol, but here the data fits CH₃O better.

Wait — CH₃O has odd number of hydrogens, which is unusual for organic compounds, but mathematically it's correct.

Perhaps it's formaldehyde CH₂O, but let's check.

For CH₂O: C=12, H=2.016, O=16, total 30.016

H% = 2.016/30.016≈6.72% — but here H% is 9.75% — too low.

For CH₄O: H%=4.032/32.042≈12.58% — too high.

So only CH₃O fits.

Empirical formula: CH₃O

In the image, they have CH₃O boxed — correct.

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#### Part e:
Given:
- A 2.50 g sample of uranium was heated in air. The resulting oxide weighed 2.949 g. Determine the empirical formula of the oxide.

Mass of U = 2.50 g
Mass of oxide = 2.949 g
So mass of O = 2.949 - 2.50 = 0.449 g

Moles:
U: 2.50 / 238.03 ≈ 0.010502 mol
O: 0.449 / 16.00 = 0.0280625 mol

Divide by smallest (0.010502):
U: 0.010502 / 0.010502 = 1
O: 0.0280625 / 0.010502 ≈ 2.672

Not integer. Multiply by 3:
U: 3
O: 8.016 → 8

So U₃O₈

Check: U₃O₈ molar mass = 3*238.03 + 8*16 = 714.09 + 128 = 842.09 g/mol

U% = 714.09/842.09 ≈ 84.8%
In sample: U mass 2.50g in 2.949g → 2.50/2.949 ≈ 84.77% — matches.

O% = 128/842.09 ≈ 15.2%, sample O=0.449/2.949≈15.23% — good.

So empirical formula: U₃O₈

In the image, they have U₃O₈ boxed — correct.

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#### Part f:
Given:
- When 2.50 g of copper reacts with oxygen, the copper oxide product has a mass of 2.81 g. What is the empirical formula of the copper oxide?

Mass Cu = 2.50 g
Mass oxide = 2.81 g
Mass O = 2.81 - 2.50 = 0.31 g

Moles:
Cu: 2.50 / 63.55 ≈ 0.03934 mol
O: 0.31 / 16.00 = 0.019375 mol

Divide by smallest (0.019375):
Cu: 0.03934 / 0.019375 ≈ 2.030
O: 0.019375 / 0.019375 = 1

So approximately 2:1 → Cu₂O

Check: Cu₂O molar mass = 2*63.55 + 16 = 127.1 + 16 = 143.1 g/mol

Cu% = 127.1/143.1 ≈ 88.82%
Sample: 2.50/2.81 ≈ 88.97% — close.

O% = 16/143.1 ≈ 11.18%, sample O=0.31/2.81≈11.03% — slight discrepancy due to rounding.

With exact:
Cu moles: 2.50 / 63.546 = 0.03934 (using 63.546)
O moles: 0.31 / 16 = 0.019375
Ratio Cu/O = 0.03934 / 0.019375 = 2.0304 — very close to 2.

So empirical formula: Cu₂O

In the image, they have Cu₂O boxed — correct.

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Now, summarizing all parts:

a. C₂H₆O (but image has C₄H₁₀O₂ — probably mistake; should be C₂H₆O)
b. AgNO₃ (assuming S was typo for N)
c. SO₂
d. CH₃O
e. U₃O₈
f. Cu₂O

But for part a, let's confirm once more.

In part a, masses: C 0.783g, H 0.196g, O 0.521g

Moles:
C: 0.783/12.01 = 0.065195
H: 0.196/1.008 = 0.194444
O: 0.521/16.00 = 0.0325625

Divide by O moles:
C: 0.065195 / 0.0325625 = 2.002
H: 0.194444 / 0.0325625 = 5.971
O: 1

So C₂H₆O — yes.

If they want molecular, but question says empirical.

In the image, for part a, they have calculations leading to C₂H₆O but then boxed C₄H₁₀O₂ — perhaps they thought it was molecular, but the problem doesn't give molar mass, so it must be empirical.

I think for accuracy, we should put C₂H₆O.

But let's see the other parts — in part d, they have CH₃O, which is correct.

So for final answers:

a. C₂H₆O
b. AgNO₃
c. SO₂
d. CH₃O
e. U₃O₈
f. Cu₂O

Now, to box them as per the format.

Final Answer:
a. C₂H₆O
b. AgNO₃
c. SO₂
d. CH₃O
e. U₃O₈
f. Cu₂O