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 table with compound (a, b, c) and mass of nitrogen that combines with 1.00 g of oxygen (1.750 g, 0.8750 g, 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 table with compound (a, b, c) and mass of nitrogen that combines with 1.00 g of oxygen (1.750 g, 0.8750 g, 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

### Part 6a
# Explanation:
## Step1: Calculate ratio for Compound I
Ratio (O:H) for Compound I: $\frac{120.0\ 	ext{g O}}{15.0\ 	ext{g H}} = 8.0$
## Step2: Calculate ratio for Compound II
Ratio (O:H) for Compound II: $\frac{32.0\ 	ext{g O}}{2.0\ 	ext{g H}} = 16.0$
# Answer:
Compound I: $8.0$, Compound II: $16.0$

### Part 6b
# Brief Explanations:
Compounds differ because the ratio of oxygen to hydrogen (mass) is different. Different mass ratios imply different atom combinations (formulas), so they are distinct compounds (e.g., water vs. hydrogen peroxide - like ratios here).
# Answer:
Different O:H mass ratios mean different atom combinations (formulas), so they are distinct compounds.

### Part 6c
# Brief Explanations:
These mass ratios are simple whole - number multiples of each other ($16.0\div8.0 = 2$). This relates to the Law of Multiple Proportions, showing that when two elements form multiple compounds, the mass ratios of one element (per fixed mass of the other) are in simple whole - number ratios.
# Answer:
The ratios are simple whole - number multiples (16.0 is 2×8.0), relating to the Law of Multiple Proportions.

### Part 6d
# Explanation:
## Step1: Recall water's formula and ratio
Water ($\ce{H2O}$) has O:H mass ratio $\frac{16\ 	ext{g O}}{2\ 	ext{g H}} = 8$ (approx, using atomic masses: O = 16, H = 1). Compound II has O:H ratio 16, which is double water's ratio.
## Step2: Deduce formula
If water has 1 O per 2 H, a ratio double that (O:H mass ratio 16) would imply 2 O per 2 H (or 1 O per 1 H in simplest terms). So the formula could be $\ce{H2O2}$ (hydrogen peroxide - like), as the O:H mass ratio for $\ce{H2O2}$ is $\frac{32\ 	ext{g O}}{2\ 	ext{g H}} = 16$, matching Compound II's ratio.
# Answer:
$\ce{H2O2}$ (Hydrogen Peroxide)

### Part 7a
# Brief Explanations:
- Compound B: Formula $\ce{NO}$. So a particle diagram for B has 1 N atom and 1 O atom bonded (e.g., a molecule with one N and one O).
- Compound A: Mass of N per 1 g O is 1.750 g. For B, it's 0.8750 g. So A has double the N per O as B. So formula for A: If B is $\ce{NO}$, A should have 2 N per 1 O? Wait, no - mass ratio: N (A)/N (B) = 1.750/0.8750 = 2. So if B is $\ce{NO}$ (1 N, 1 O), A has 2 N per 1 O? Wait, no, mass of N: atomic mass N = 14, O = 16. For B ($\ce{NO}$), mass N per 1 g O: $\frac{14\ 	ext{g N}}{16\ 	ext{g O}}\approx0.875\ 	ext{g N/g O}$ (matches B's data). For A, 1.750 g N/g O is double B's. So A has 2 N per 1 O? Wait, no, per 1 g O, A has 2× the N of B. So formula for A: $\ce{N2O}$ (2 N, 1 O). Let's check: mass N per 1 g O: $\frac{2	imes14\ 	ext{g N}}{16\ 	ext{g O}}=\frac{28}{16}=1.75\ 	ext{g N/g O}$ (matches A's data).
- Compound C: Mass of N per 1 g O is 0.4375 g, which is half of B's (0.8750/2 = 0.4375). So C has half the N per O as B. So formula for C: If B is $\ce{NO}$ (1 N, 1 O), C has 0.5 N per 1 O? Wait, atomic mass - based: For C, mass N per 1 g O is 0.4375. B is 0.875. So C has 1/2 the N of B per O. So formula: $\ce{NO2}$? Wait, no. Wait, B is $\ce{NO}$ (1 N, 1 O). C: mass N per 1 g O is 0.4375, which is 1/2 of B's 0.875. So number of N atoms: if B has 1 N, C has 0.5 N? No, we need whole atoms. Wait, let's use the mass ratio. N (C)/N (B) = 0.4375/0.8750 = 0.5 = 1/2. So if B has 1 N per 1 O, C has 1 N per 2 O (since per 2 O, N would be 0.875 g, but per 1 O, it's 0.4375 g). So formula $\ce{NO2}$: mass N per 1 g O: $\frac{14\ 	ext{g N}}{32\ 	ext{g O}} = 0.4375\ 	ext{g N/g O}$ (matches C's data).

So particle diagrams:
- A ($\ce{N2O}$): Molecule with 2 N atoms and 1 O atom.
- B ($\ce{NO}$): Molecule with 1 N atom and 1 O atom.
- C ($\ce{NO2}$): Molecule with 1 N atom and 2 O atoms.
# Answer:
- A: $\ce{N2O}$ (diagram: 2 N - 1 O bonded)
- B: $\ce{NO}$ (diagram: 1 N - 1 O bonded)
- C: $\ce{NO2}$ (diagram: 1 N - 2 O bonded)

### Part 7b
# Brief Explanations:
- For B ($\ce{NO}$), mass N per 1 g O: $\frac{14}{16}=0.875\ 	ext{g}$ (matches data).
- For A ($\ce{N2O}$), mass N per 1 g O: $\frac{2	imes14}{16}=1.75\ 	ext{g}$ (matches A's 1.750 g).
- For C ($\ce{NO2}$), mass N per 1 g O: $\frac{14}{32}=0.4375\ 	ext{g}$ (matches C's 0.4375 g). The mass ratios of N (per 1 g O) are 1.75:0.875:0.4375 = 4:2:1 (or 2:1:0.5, but simplified whole - number ratios: A:B:C = 2:1:0.5 → multiply by 2: 4:2:1, but more simply, A has double N of B, C has half N of B). This follows the Law of Multiple Proportions, justifying the formulas (ratios of N atoms per O atom are whole - number multiples: 2, 1, 0.5 → but as whole numbers for atoms, 2,1, and 1 with double O for C).
# Answer:
The mass ratios of N (per 1 g O) are whole - number multiples (A:B = 2:1, B:C = 2:1), matching the atom ratios in formulas ($\ce{N2O}$:$\ce{NO}$:$\ce{NO2}$ have N:O ratios 2:1, 1:1, 1:2), justifying the representations.

### Part 8
# Brief Explanations:
- In Question 6: Compounds of H and O. Compound I (water - like) has O:H ratio 8, Compound II has ratio 16. 16 is 2×8, a simple whole - number multiple.
- In Question 7: Compounds of N and O. Mass of N per 1 g O: A (1.75), B (0.875), C (0.4375). Ratios: A:B = 2:1, B:C = 2:1, A:C = 4:1 (all simple whole - number multiples). The Law of Multiple Proportions states that when two elements form multiple compounds, the masses of one element that combine with a fixed mass of the other are in simple whole - number ratios. Both questions show this: fixed H (or O) mass, O (or N) masses are in whole - number ratios.
# Answer:
In Q6, O:H mass ratios (8, 16) are 2× multiple. In Q7, N:O mass ratios (per 1g O) are 2:1 (A:B), 2:1 (B:C) - simple whole - number ratios. Both match the Law of Multiple Proportions (masses of one element per fixed mass of another are in whole - number ratios for multiple compounds).

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QUESTION IMAGE

Question

Explanation:

Part 6a

Step1: Calculate ratio for Compound I

Step2: Calculate ratio for Compound II

Answer:

Part 6b