Question 1:

To determine the empirical formula of a compound containing 79.9% carbon and 20.1% hydrogen by mass, follow these steps:

1. Assume a 100 g sample:

   • Mass of carbon = 79.9 g

   • Mass of hydrogen = 20.1 g

2. Convert masses to moles:

   • Moles of carbon = 79.9/12.01≈6.6536 mol

   • Moles of hydrogen = 20.1/1.008≈19.9405 mol

3. Find the simplest mole ratio:

   • Divide each by the smallest number of moles (which is carbon, 6.6536):

     • Carbon: 6.6536 / 6.6536=1

     • Hydrogen: 19.9405 / 6.6536≈2.996≈3

4. Write the empirical formula:

   • The ratio is C : H = 1 : 3, so the empirical formula is CH₃.

Answer: B. CH₃

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 2:

Assume a 100 g sample of the oxide:

• Mass of oxygen = 32 g

• Mass of metal X = 100 g - 32 g = 68 g

Now, convert masses to moles:

• Moles of X = mass / atomic mass = 68 g / 51 g/mol ≈ 1.3333 mol

• Moles of O = 32 g / 16 g/mol = 2.000 mol

Find the simplest mole ratio by dividing both by the smaller number of moles (which is 1.3333):

• For X: 1.3333 / 1.3333 = 1

• For O: 2.000 / 1.3333 ≈ 1.5

To get whole numbers, multiply both by 2:

• X: 1 × 2 = 2

• O: 1.5 × 2 = 3

Thus, the empirical formula is X₂O₃.

Answer: D. X₂O₃

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 3:

To determine the empirical formula of the oxide of sulfur given:

• Mass of sulfur = 1.6 g

• Mass of oxygen = 2.4 g

Follow these steps:

1. Convert masses to moles:

   • Molar mass of sulfur (S) = 32 g/mol

   • Molar mass of oxygen (O) = 16 g/mol

   • Moles of S = 1.6/32=0.05 mol

   • Moles of O = 2.4/16=0.15 mol

2. Find the simplest mole ratio:

   • Divide both by the smallest number of moles (which is 0.05):

     • S: 0.05/0.05=1

     • O: 0.15/0.05=3

3. Write the empirical formula:

   • The ratio is S : O = 1 : 3, so the empirical formula is SO₃.

Answer: B. SO₃

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 4:

To find the molecular formula of melamine given:

• Empirical formula = CH₂N₂

• Molecular mass = 126 g/mol

Follow these steps:

1. Calculate the empirical formula mass:

   • Carbon (C): 12 g/mol × 1 = 12 g/mol

   • Hydrogen (H): 1 g/mol × 2 = 2 g/mol

   • Nitrogen (N): 14 g/mol × 2 = 28 g/mol

   • Total empirical mass = 12 + 2 + 28 = 42 g/mol

2. Determine the multiple (n):

   • n = Molecular mass / Empirical mass = 126 / 42 = 3

3. Multiply the empirical formula by n:

   • Molecular formula = (CH₂N₂) × 3 = C₃H₆N₆

Answer: C. C₃H₆N₆

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 5:

To determine the empirical formula of the oxide of chromium:

Given:

• Mass of the oxide = 5.0 g

• Mass of chromium in the oxide = 2.6 g

First, find the mass of oxygen:

• Mass of oxygen = Total mass of oxide - Mass of chromium = 5.0 g - 2.6 g = 2.4 g

Now, convert masses to moles:

• Molar mass of chromium (Cr) = 52 g/mol

• Molar mass of oxygen (O) = 16 g/mol

• Moles of Cr = 2.6/52=0.05 mol

• Moles of O = 2.4/16=0.15 mol

Find the simplest mole ratio by dividing both by the smallest number of moles (0.05):

• Cr: 0.05/0.05=1

• O: 0.15/0.05=3

The ratio is Cr : O = 1 : 3, so the empirical formula is CrO₃.

Answer: B. CrO₃

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 6:

To determine which compounds have the empirical formula CH₂O, we need to simplify each molecular formula to its simplest whole-number ratio.

Recall: The empirical formula CH₂O has a ratio of C : H : O = 1 : 2 : 1.

Let's analyze each compound:

I. C₆H₁₂O₆

• Divide all subscripts by 6: C₆/₆ H₁₂/₆ O₆/₆ = CH₂O

• So, it has the empirical formula CH₂O.

II. CH₃COOH (which is C₂H₄O₂)

• Divide all subscripts by 2: C₂/₂ H₄/₂ O₂/₂ = CH₂O

• So, it has the empirical formula CH₂O.

III. C₁₂H₂₄O₁₁

• The ratio is C : H : O = 12 : 24 : 11.

• This cannot be simplified to a 1:2:1 ratio because 11 (for oxygen) is not divisible to give 1. The closest would be C₁₂H₂₄O₁₁, which is not CH₂O (since CH₂O requires O=1, but here O=11/12? Not integer).

• Actually, divide by 12: C₁₂/₁₂ H₂₄/₁₂ O₁₁/₁₂ = C₁ H₂ O₀.9167... which is not a whole number. So, it does not simplify to CH₂O.

Therefore, only I and II have the empirical formula CH₂O.

Answer: B. I and II only

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 7:

To determine the molecular formula of the hydrocarbon, let's analyze the combustion reaction:

The hydrocarbon (CxHy) burns in oxygen to produce CO₂ and H₂O:

Given:

• 0.25 mol of hydrocarbon produces:

  • 44.0 g of CO₂

  • 22.53 g of H₂O

Step 1: Find the moles of CO₂ and H₂O produced

• Molar mass of CO₂ = 44 g/mol

• Moles of CO₂ = 44.0/44=1.0 mol

• Molar mass of H₂O = 18 g/mol

• Moles of H₂O = 22.53/18≈1.2517 mol

Step 2: Relate these to the moles of hydrocarbon
Since 0.25 mol of hydrocarbon produces:

• 1.0 mol of CO₂ ⇒ This means x=(moles of CO₂) / (moles of hydrocarbon)
  x =1.0/0.25=4
  

• 1.2517 mol of H₂O ⇒ This means (y/2)=(moles of H₂O) / (moles of hydrocarbon) (y/2)=1.2517/0.25=5.0068≈5
  So, y=10

Step 3: Determine the molecular formula
The hydrocarbon has the formula C₄H₁₀.

Answer: C. C₄H₁₀

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 9:

To determine the empirical formula of the hydrocarbon from its combustion products:

Given:

• Mass of CO₂ produced = 4.40 g

• Mass of H₂O produced = 2.70 g

Step 1: Find the moles of carbon and hydrogen in the hydrocarbon

• All carbon in the hydrocarbon ends up in CO₂.

• Molar mass of CO₂ = 44 g/mol

• Moles of CO₂ = 4.40/44=0.10 mol

• Each mole of CO₂ contains 1 mole of C, so moles of C = 0.10 mol

• All hydrogen in the hydrocarbon ends up in H₂O.

• Molar mass of H₂O = 18 g/mol

• Moles of H₂O = 2.70/18=0.15 mol

• Each mole of H₂O contains 2 moles of H, so moles of H = 0.15 × 2 = 0.30 mol

Step 2: Find the simplest mole ratio (C : H)

• Divide both by the smallest number (0.10):

  • C: 0.10/0.10=1

  • H: 0.30/0.10=3

So, the empirical formula is CH₃.

Answer: C. CH₃

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 10:

To determine which compound has the highest percentage by mass of nitrogen, we'll calculate the mass percentage of nitrogen in each option.

A. NH₃ (Ammonia)

• Molar mass = 14 + 3(1) = 17 g/mol

• Mass of N = 14 g

• % N = 14/17×100%≈82.35%

B. NH₄NO₃ (Ammonium nitrate)

• Molar mass = 14 + 4(1) + 14 + 3(16) = 18 + 14 + 48 = 80 g/mol

• Mass of N = 14 + 14 = 28 g (two N atoms)

• % N = 28/80×100%=35.00%

C. CO(NH₂)₂ (Urea)

• Molar mass = 12 + 16 + 2(14) + 4(1) = 12 + 16 + 28 + 4 = 60 g/mol

• Mass of N = 28 g (two N atoms)

• % N = 28/60×100%≈46.67%

D. (NH₄)₂SO₄ (Ammonium sulfate)

• Molar mass = 2(14) + 8(1) + 32 + 4(16) = 28 + 8 + 32 + 64 = 132 g/mol

• Mass of N = 28 g (two N atoms)

• % N = 28/132×100%≈21.21%

Comparing the percentages:

• A. NH₃: ~82.35%

• B. NH₄NO₃: 35.00%

• C. Urea: ~46.67%

• D. (NH₄)₂SO₄: ~21.21%

NH₃ has the highest percentage by mass of nitrogen.

Answer: A. NH₃

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.