Question

A 75 gram sample of a white powder is chemically broken down into 29.86 grams of copper, 15.06 grams of sulfur, and an unknown amount of oxygen gas. How much oxygen gas was made?

Answer 1

Suppose you want to make 11.2 grams of the white powder with no leftovers, how much copper, sulfur, and oxygen would you have to use?

Answer 2

To find the mass of oxygen gas produced, we can start by using the law of conservation of mass, which states that the total mass of the reactants must equal the total mass of the products.

We know the total mass of the sample and the masses of copper and sulfur:

- Total mass of the sample = 75 grams
- Mass of copper = 29.86 grams
- Mass of sulfur = 15.06 grams

Let's denote the mass of oxygen gas produced as \( m_{\text{oxygen}} \).

According to the conservation of mass, we can write the equation:

\[
m_{\text{sample}} = m_{\text{copper}} + m_{\text{sulfur}} + m_{\text{oxygen}}
\]

Substituting the known values into the equation:

\[
75 \, \text{grams} = 29.86 \, \text{grams} + 15.06 \, \text{grams} + m_{\text{oxygen}}
\]

Now we can combine the masses of copper and sulfur:

\[
75 \, \text{grams} = 44.92 \, \text{grams} + m_{\text{oxygen}}
\]

Next, we solve for \( m_{\text{oxygen}} \):

\[
m_{\text{oxygen}} = 75 \, \text{grams} - 44.92 \, \text{grams}
\]

\[
m_{\text{oxygen}} = 30.08 \, \text{grams}
\]

Thus, the mass of oxygen gas produced is **30.08 grams**.

Answer 3

To determine how much copper, sulfur, and oxygen are needed to create 11.2 grams of the white powder without leftovers, we will first establish the mass ratio between the elements based on the original sample. We know the composition from the initial breakdown:

From the original sample of 75 grams:
- Mass of copper: 29.86 grams
- Mass of sulfur: 15.06 grams
- Mass of oxygen: 30.08 grams (as calculated previously)

Now, we can find the proportions of each component in the original sample:

1. **Total mass of sample**:
\[
\text{Total} = 29.86 \, \text{g} + 15.06 \, \text{g} + 30.08 \, \text{g} = 75 \, \text{g}
\]

2. **Proportions**:
- Proportion of copper:
\[
\text{Proportion of copper} = \frac{29.86}{75}
\]
- Proportion of sulfur:
\[
\text{Proportion of sulfur} = \frac{15.06}{75}
\]
- Proportion of oxygen:
\[
\text{Proportion of oxygen} = \frac{30.08}{75}
\]

Now, let's calculate these proportions:

\[
\text{Proportion of copper} \approx 0.3971
\]
\[
\text{Proportion of sulfur} \approx 0.2008
\]
\[
\text{Proportion of oxygen} \approx 0.4001
\]

Next, we apply these proportions to the new total mass of the white powder (11.2 grams):

1. **Mass of copper**:
\[
\text{Mass of copper} = 11.2 \, \text{g} \times 0.3971 \approx 4.44 \, \text{g}
\]

2. **Mass of sulfur**:
\[
\text{Mass of sulfur} = 11.2 \, \text{g} \times 0.2008 \approx 2.25 \, \text{g}
\]

3. **Mass of oxygen**:
\[
\text{Mass of oxygen} = 11.2 \, \text{g} \times 0.4001 \approx 4.48 \, \text{g}
\]

In summary, to create 11.2 grams of the white powder without leftovers, you would need approximately:

- **Copper**: 4.44 grams
- **Sulfur**: 2.25 grams
- **Oxygen**: 4.48 grams

If you prefer rounding to two decimal places, you could state the required amounts as:

- Copper: **4.44 grams**
- Sulfur: **2.25 grams**
- Oxygen: **4.48 grams**