6. two compounds of hydrogen and oxygen are tested. compound i contains 15.0 g of hydrogen and 120.0 g of oxygen. compound ii contains 2.0 g of hydrogen and 32.0 g of oxygen.
a. determine the ratio of the mass of oxygen to the mass of hydrogen in each of the compounds.
b. why are the compounds not the same?
c. what is significant about these mass ratios?
d. if compound i is water, what could be the formula of compound ii?
7. nitrogen and oxygen combine to form a variety of compounds. the following data were collected for three different compounds of nitrogen and oxygen:
analysis data of nitrogen & oxygen compounds
compound | mass of nitrogen that combines with 1.00 g of oxygen
a | 1.750 g
b | 0.8750 g
c | 0.4375 g
a. additional evidence shows that the formula of compound b is no. sketch particle diagrams of molecules of all three compounds.
b. justify your representations above.
8. explain how the examples in questions 6 and 7 help to illustrate the law of multiple proportions.

Question

6. two compounds of hydrogen and oxygen are tested. compound i contains 15.0 g of hydrogen and 120.0 g of oxygen. compound ii contains 2.0 g of hydrogen and 32.0 g of oxygen.
  a. determine the ratio of the mass of oxygen to the mass of hydrogen in each of the compounds.
  b. why are the compounds not the same?
  c. what is significant about these mass ratios?
  d. if compound i is water, what could be the formula of compound ii?
7. nitrogen and oxygen combine to form a variety of compounds. the following data were collected for three different compounds of nitrogen and oxygen:
  analysis data of nitrogen & oxygen compounds
  compound | mass of nitrogen that combines with 1.00 g of oxygen
  a | 1.750 g
  b | 0.8750 g
  c | 0.4375 g
  a. additional evidence shows that the formula of compound b is no. sketch particle diagrams of molecules of all three compounds.
  b. justify your representations above.
8. explain how the examples in questions 6 and 7 help to illustrate the law of multiple proportions.

Accepted Answer

### Question 6a
# Explanation:
## Step1: Calculate ratio for Compound I
Mass of oxygen = 120.0 g, Mass of hydrogen = 15.0 g. Ratio (O:H) = $\frac{120.0}{15.0}$ = 8.
## Step2: Calculate ratio for Compound II
Mass of oxygen = 32.0 g, Mass of hydrogen = 2.0 g. Ratio (O:H) = $\frac{32.0}{2.0}$ = 16.
# Answer:
Compound I: 8:1, Compound II: 16:1

### Question 6b
# Brief Explanations:
Compounds have different O:H mass ratios, meaning different atom combinations (formulas), so they are different substances (e.g., water vs. hydrogen peroxide).
# Answer:
Different oxygen - to - hydrogen mass ratios (different chemical formulas)

### Question 6c
# Brief Explanations:
Ratios are simple whole numbers, consistent with the Law of Multiple Proportions (atoms combine in whole - number ratios), indicating different compounds of H and O.
# Answer:
Ratios are simple whole numbers (8 and 16, or 1:2 when compared), showing different compounds of H and O form with whole - number atom ratios.

### Question 6d
# Explanation:
## Step1: Recall water's formula (H₂O)
In water (H₂O), O:H mass ratio is 8:1 (16 g O : 2 g H ≈ 8:1). Compound II has O:H ratio 16:1 (32 g O : 2 g H).
## Step2: Determine formula
Let the formula be HₓOᵧ. Mass ratio O:H = $\frac{16y}{x}$ = 16 (since atomic mass O = 16, H = 1). So $\frac{16y}{x}$ = 16 → y = x. A simple formula with x = y = 2? Wait, no: For H₂O₂ (hydrogen peroxide), O:H mass ratio: (2×16):(2×1) = 32:2 = 16:1. So formula is H₂O₂.
# Answer:
H₂O₂ (Hydrogen Peroxide)

### Question 7a
# Brief Explanations:
- Compound B: NO (1 N, 1 O). For 1 g O, mass of N in B is 0.8750 g.
- Compound A: Mass of N per 1 g O is 1.750 g, which is 2×0.8750 g. So ratio of N in A:B is 2:1. So formula: N₂O (2 N, 1 O).
- Compound C: Mass of N per 1 g O is 0.4375 g, which is 0.5×0.8750 g. Ratio of N in C:B is 1:2. So formula: NO₂ (1 N, 2 O).

Particle diagrams:
- Compound A (N₂O): Two N atoms bonded to one O atom.
- Compound B (NO): One N atom bonded to one O atom.
- Compound C (NO₂): One N atom bonded to two O atoms.
# Answer:
Compound A: N₂O (diagram: two N spheres and one O sphere bonded), Compound B: NO (one N and one O sphere bonded), Compound C: NO₂ (one N and two O spheres bonded)

### Question 7b
# Brief Explanations:
- For 1 g O, N masses: A (1.750 g), B (0.8750 g), C (0.4375 g). Ratios A:B:C = 1.750:0.8750:0.4375 = 4:2:1 (or 2:1:0.5, but whole - number ratio 4:2:1).
- Compound B: NO (1 N, 1 O). Compound A: N₂O (2 N, 1 O) (since 2×N in B for same O). Compound C: NO₂ (1 N, 2 O) (since 0.5×N in B for same O, so O is doubled). These formulas have whole - number atom ratios, justifying the mass ratios.
# Answer:
Mass ratios of N to 1 g O (1.750:0.8750:0.4375) simplify to 4:2:1 (or 2:1:0.5, whole - number 4:2:1). Compound B (NO) has 1 N:1 O. Compound A (N₂O) has 2 N:1 O (2×N in B for same O), Compound C (NO₂) has 1 N:2 O (O doubled from B). So atom ratios are whole numbers, matching mass ratios.

### Question 8
# Brief Explanations:
- Law of Multiple Proportions: When two elements form multiple compounds, the mass ratios of one element (for a fixed mass of the other) are simple whole numbers.
- Question 6: H and O form two compounds. O mass for fixed H (2 g H): Compound I (32 g O? No, wait 15 g H - 120 g O; 2 g H would have (120/15)×2 = 16 g O? Wait no, earlier calculation: Compound I (H₂O): O:H = 8:1 (16 g O : 2 g H). Compound II (H₂O₂): O:H = 16:1 (32 g O : 2 g H). Ratio of O masses (for 2 g H) is 16:32 = 1:2 (simple whole number).
- Question 7: N and O form three compounds. For 1 g O, N masses: A (1.750), B (0.8750), C (0.4375). Ratios A:B = 2:1, B:C = 2:1, A:C = 4:1 (all simple whole numbers). So both questions show that for fixed mass of one element, the mass of the other element in different compounds is in simple whole - number ratios, illustrating the Law.
# Answer:
In Q6, H and O form 2 compounds. For fixed H (2 g), O masses are 16 g (H₂O) and 32 g (H₂O₂), ratio 1:2 (simple whole number). In Q7, N and O form 3 compounds. For 1 g O, N masses have ratios 2:1 (A:B), 2:1 (B:C), 4:1 (A:C) (simple whole numbers). Both follow the Law of Multiple Proportions: when two elements form multiple compounds, the mass of one element (per fixed mass of the other) is in simple whole - number ratios.

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QUESTION IMAGE

Question

Explanation:

Question 6a

Step1: Calculate ratio for Compound I

Step2: Calculate ratio for Compound II

Answer:

Question 6b