Question

You have gone back in time and are working with Dalton on a table of relative masses. Following are his data:

0.602g gas A reacts with 0.295g gasB

0.172 g gas B reacts with 0.401 g gasC

0.320 g gas A reacts with 0.374g gas C

a. Assuming simplest formulas (AB,BC, and AC), construct a table of relative masses for Dalton.

b. Knowing some history of chemistry, you tell Dalton that if he determines the volumes of the gases reacted at constant temperature and pressure, he need not assume simplest formulas. You collect the following data:

$\mathbf{6}\mathbf{}\mathbf{volumes}\mathbf{}\mathbf{gas}\mathbf{}\mathbf{A}\mathbf{+}\mathbf{1}\mathbf{}\mathbf{volume}\mathbf{}\mathbf{gas}\mathbf{}\mathbf{B}\mathbf{\to }\mathbf{4}\mathbf{}\mathbf{volumes}\mathbf{}\mathbf{product}$

$\mathbf{1}\mathbf{volumes}\mathbf{}\mathbf{gasB}\mathbf{+}\mathbf{4}\mathbf{}\mathbf{volume}\mathbf{}\mathbf{gasC}\mathbf{\to }\mathbf{4}\mathbf{volumes}\mathbf{}\mathbf{product}$

$\mathbf{3}\mathbf{volumes}\mathbf{}\mathbf{gasA}\mathbf{+}\mathbf{3}\mathbf{volumes}\mathbf{}\mathbf{gasC}\mathbf{\to }\mathbf{6}\mathbf{volumes}\mathbf{}\mathbf{product}$

Write the simplest balanced equations, and find the actual relative masses of the elements.

Explaining your reasoning.

Short answer

a. Relative mass tableassuming simplest formulas (AB, BC, and AC) is:

A	B	C
6.0	3.0	7.0
6.0	9.0	7.0

b. The set of balanced reactionsthat confirms the initial mass data is correct.

Step-by-step solution

Definition of Relative mass

For each set of data, divide the larger number by the smaller number to determine relative masses.

$\frac{0.602}{0.295}=2.04;\mathrm{A}=2.04\mathrm{when}\mathrm{B}=1.00$

$\frac{0.401}{0.172}=2.33;\mathrm{A}=2.33\mathrm{when}\mathrm{B}=1.00$

$\frac{0.374}{0.320}=1.17;\mathrm{A}=1.17\mathrm{when}\mathrm{B}=1.00$

$\frac{0.374}{0.320}=1.17;\mathrm{A}=1.17\mathrm{when}\mathrm{B}=1.00$

To have whole numbers, multiply the results by 3. We can have a table:

A	B	C
6.0	3.0	7.0
6.0	9.0	7.0

Assuming 6.0 for the relative mass of A, the relative masses would be A=6.0, B=3.0, and C=7.0 (if simplest formulas are assumed).

Explanation

Gas volumes are proportional to the number of molecules present. There are many possible correct answers for the balanced equations. One such solution that fits the gas volume data is:

$$\begin{gathered} 6{\mathrm{A}}_2+{\mathrm{B}}_4\to 4{\mathrm{A}}_3\mathrm{B} \\ {\mathrm{B}}_4+4\mathrm{C}+3\to 4{\mathrm{BC}}_3 \\ 3{\mathrm{A}}_2+2{\mathrm{C}}_3\to 6\mathrm{AC} \end{gathered}$$

In any correct set of reactions, the calculated mass data must match the mass data given initially in the problem. Here the new table of relative masses would be 3pt.

$$\begin{gathered} \frac{6\times {\mathrm{massA}}_2}{{\mathrm{massB}}_4}=\frac{0.602}{0.295} \\ {\mathrm{massA}}_2=0.340\left[{\mathrm{massB}}_4\right]6\mathrm{pt} \end{gathered}$$

$$\begin{gathered} \frac{4\times {\mathrm{massC}}_3}{{\mathrm{massB}}_4}=\frac{0.401}{0.172} \\ {\mathrm{massC}}_3=0.583\left[{\mathrm{massB}}_4\right]6\mathrm{pt} \end{gathered}$$

$$\begin{gathered} \frac{2\times {\mathrm{massC}}_3}{{\mathrm{massB}}_4}=\frac{0.374}{0.320} \\ {\mathrm{massA}}_2=0.570\left[{\mathrm{massC}}_3\right]3\mathrm{pt} \end{gathered}$$

Assume some relative mass number for any of the masses, we will assume that mass B=3.0, so mass${\mathrm{B}}_4=4\times 3.0=12$.

Mass calculation

$$\begin{gathered} \mathrm{Mass}{\mathrm{C}}_3=0.583\times 12=7.0,\mathrm{mass}\mathrm{C}=7.0/3 \\ \mathrm{Mass}{\mathrm{A}}_2=0.570\times 7=4.0,\mathrm{mass}\mathrm{A}=4.0/2=2.0 \end{gathered}$$

Conclusion

Note that any set of balanced reactions that confirms the initial mass data is correct. This is just one possibility.