Question

note: you may write your steps to calculation on separate piece of paper
table 1—data collection
unknown sample | number of moles | mass of empty container | mass of chemical with container | mass of chemical substances
--- | --- | --- | --- | ---
0 (example) | 0.044986 | 6.38 g | 11.76g | 11.76 - 6.38 = 5.38 g
1a | 0.055172 | 7.35 g | 10.66 | 3.31g
1b | 0.108620 | 7.50 | 13.89 | 6.39
2a | 0.028378 | 0.85g | 1.11g | (1.11g - 0.85g = 0.26g)
2b | 0.041892 | 7.1g | 10.3g |
3a | 0.044366 | 7.35g | 13.76g |
3b | 0.038028 | 7.5g | 12.3b |
4a | 0.041509 | 7.4g | 12.6g |
4b | 0.051887 | 7.4g | 14g |
5a | 0.093504 | 7.1g | 10.3g |
5b | 0.030326 | 7.87g | 18.57g |
6a | 0.031111 | 7.4g | 13.12g |
6b | 0.045555 | 7.37g | 15.62g |
7a | 0.107692 | 7.25g | 14.48g |
7b | 0.355384 | 7.4g | 30.5g |

Answer:

To solve for the mass of chemical substances for each unknown sample, we use the formula:

$\text{Mass of Chemical Substances} = \text{Mass of Chemical with container} - \text{Mass of empty container}$

Example (Sample 0):

Given:

• Mass of empty container = $6.38\ \text{g}$
• Mass of chemical with container = $11.76\ \text{g}$

$\text{Mass of Chemical Substances} = 11.76 - 6.38 = 5.38\ \text{g}$

Sample 1A:

Given:

• Mass of empty container = $7.35\ \text{g}$ (from the table)
• Mass of chemical with container = $10.66\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 10.66 - 7.35 = 3.31\ \text{g}$

Sample 1B:

Given:

• Mass of empty container = $7.50\ \text{g}$ (from the table)
• Mass of chemical with container = $13.89\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 13.89 - 7.50 = 6.39\ \text{g}$

Sample 2A:

Given:

• Mass of empty container = $0.85\ \text{g}$ (from the table)
• Mass of chemical with container = $1.11\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 1.11 - 0.85 = 0.26\ \text{g}$

Sample 2B:

Given:

• Mass of empty container = $7.10\ \text{g}$ (from the table)
• Mass of chemical with container = $10.39\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 10.39 - 7.10 = 3.29\ \text{g}$ (Note: The table’s handwritten $1.11g$ might be a typo; assuming $10.39g$ for “mass with container.”)

Sample 3A:

Given:

• Mass of empty container = $7.35\ \text{g}$ (from the table)
• Mass of chemical with container = $13.76\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 13.76 - 7.35 = 6.41\ \text{g}$

Sample 3B:

Given:

• Mass of empty container = $7.50\ \text{g}$ (from the table)
• Mass of chemical with container = $12.38\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 12.38 - 7.50 = 4.88\ \text{g}$

Sample 4A:

Given:

• Mass of empty container = $7.40\ \text{g}$ (from the table)
• Mass of chemical with container = $12.67\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 12.67 - 7.40 = 5.27\ \text{g}$

Sample 4B:

Given:

• Mass of empty container = $7.49\ \text{g}$ (from the table)
• Mass of chemical with container = $14.0\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 14.0 - 7.49 = 6.51\ \text{g}$

Sample 5A:

Given:

• Mass of empty container = $7.10\ \text{g}$ (from the table)
• Mass of chemical with container = $10.30\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 10.30 - 7.10 = 3.20\ \text{g}$

Sample 5B:

Given:

• Mass of empty container = $7.87\ \text{g}$ (from the table)
• Mass of chemical with container = $18.57\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 18.57 - 7.87 = 10.70\ \text{g}$

Sample 6A:

Given:

• Mass of empty container = $7.49\ \text{g}$ (from the table)
• Mass of chemical with container = $13.12\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 13.12 - 7.49 = 5.63\ \text{g}$

Sample 6B:

Given:

• Mass of empty container = $7.37\ \text{g}$ (from the table)
• Mass of chemical with container = $15.62\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 15.62 - 7.37 = 8.25\ \text{g}$

Sample 7A:

Given:

• Mass of empty container = $7.25\ \text{g}$ (from the table)
• Mass of chemical with container = $14.48\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 14.48 - 7.25 = 7.23\ \text{g}$

Sample 7B:

Given:

• Mass of empty container = $7.49\ \text{g}$ (from the table)
• Mass of chemical with container = $30.59\ \text{g}$ (from the table)

$\text{Mass of Chemical Substances} = 30.59 - 7.49 = 23.10\ \text{g}$

Final Answers (Selected Samples):

• Sample 0: $\boldsymbol{5.38\ \text{g}}$
• Sample 1A: $\boldsymbol{3.31\ \text{g}}$
• Sample 1B: $\boldsymbol{6.39\ \text{g}}$
• Sample 2A: $\boldsymbol{0.26\ \text{g}}$
• Sample 2B: $\boldsymbol{3.29\ \text{g}}$ (or as per corrected “mass with container”)

(Repeat the same process for all samples using the formula $ \text{Mass of Chemical Substances} = \text{Mass with container} - \text{Mass of empty container} $.)