1. 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 gas B
0.172g gas B reacts with 0.401g gas C
0.320g gas A reacts with 0.374g gas C
a) Assuming simplest formulas (AB,BC,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:
6 volumes gas A + 1 volume gas B -------> 4 volumes product
1 volume gas B + 4 volumes gas C --------> 4 volumes product
3 volumes gas A + 2 volumes gas C --------> 6 volumes product
Write the simplest balanced equations, and find the actual relative masses of the elements. Explain your reasoning.
a) To construct a table of relative masses for Dalton, we can analyze the given data and calculate the proportions between the masses of the gases.
From the first set of data:
0.602g of gas A reacts with 0.295g of gas B
0.172g of gas B reacts with 0.401g of gas C
0.320g of gas A reacts with 0.374g of gas C
Now, let's find the ratios between the masses:
The ratio between gas A and gas B is given by:
0.602g A / 0.295g B = 2.044
The ratio between gas B and gas C is given by:
0.172g B / 0.401g C = 0.429
The ratio between gas A and gas C is given by:
0.320g A / 0.374g C = 0.856
Therefore, the table of relative masses for Dalton would be as follows:
Gas A | Gas B | Gas C
------------------------
2.044 | 1 | 0.429
------------------------
1 | 0.429 | 0.856
------------------------
2.526 | 1.081 | 1
b) Knowing the volumes of the gases reacted at constant temperature and pressure, we can determine the balanced equations and find the actual relative masses of the elements.
Based on the given data, the balanced equations are:
6A + B --> 4 product
B + 4C --> 4 product
3A + 2C --> 6 product
From the equations, we can see that:
- For every 6 volumes of gas A, we need 1 volume of gas B.
- For every 1 volume of gas B, we need 4 volumes of gas C.
To determine the actual relative masses of the elements, we need to find the relationship between the volumes and the masses by comparing the equations.
Let's assign arbitrary values to represent the mass of gas A, B, and C:
Let's assume the mass of gas A = x g
The mass of gas B will be 2x044 times the mass of A since we know the ratio between gas A and gas B is 2.044.
Therefore, the mass of gas B = 2.044x g
Similarly, the mass of gas C will be 1.081 times the mass of B since we know the ratio between gas B and gas C is 1.081.
Therefore, the mass of gas C = 1.081 * 2.044x g
Now, let's write the equations for the masses:
6x + 2.044x = 4x (from the first equation)
1.081 * 2.044x + 4 * 1.081 * 2.044x = 4 * 1.081 * 2.044x (from the second equation)
3x + 2 * 1.081 * 2.044x = 6 * 1.081 * 2.044x (from the third equation)
Simplifying the equations:
8.044x = 4x
8.685x = 8.685x
6.831x = 6.831x
Since the coefficients on both sides of the equations are equal, the masses of the elements are the same as the coefficients. Therefore, the actual relative masses of the elements are:
Gas A: 6.831
Gas B: 8.044
Gas C: 8.685
Note: The actual relative masses are obtained by considering the volumes and applying the law of multiple proportions.
a) To construct a table of relative masses for Dalton, we need to determine the simplest formulas (molecular formulas) for gases A, B, and C. We can do this by comparing the masses of gases reacted with each other.
1. From the first reaction: 0.602g gas A reacts with 0.295g gas B
The ratio of the masses is (0.602g)/(0.295g) ≈ 2:1
This suggests that the simplest formula for gas A is AB and for gas B is B.
2. From the second reaction: 0.172g gas B reacts with 0.401g gas C
The ratio of the masses is (0.172g)/(0.401g) ≈ 1:2.33
We can multiply the ratio by 3 to get a whole number ratio: 3:7
This suggests that the simplest formula for gas C is BC.
3. From the third reaction: 0.320g gas A reacts with 0.374g gas C
The ratio of the masses is (0.320g)/(0.374g) ≈ 0.855:1
We can multiply the ratio by 2 to get a whole number ratio: 2:2
This suggests that the simplest formula for gas A is AC.
Therefore, the table of relative masses for Dalton is:
Gas A (AB) - 0.602g
Gas B (B) - 0.295g
Gas C (BC) - 0.401g
b) With the additional information on volumes of gases reacted, we can determine the simplest balanced equations and find the actual relative masses of the elements.
1. From the equation: 6 volumes gas A + 1 volume gas B -------> 4 volumes product
We can write the balanced equation as:
6AB + B ------> 4product
This equation suggests that the ratio of gas A to gas B is 6:1.
2. From the equation: 1 volume gas B + 4 volumes gas C --------> 4 volumes product
We can write the balanced equation as:
B + 4BC ------> 4product
This equation suggests that the ratio of gas B to gas C is 1:4.
3. From the equation: 3 volumes gas A + 2 volumes gas C --------> 6 volumes product
We can write the balanced equation as:
3AB + 2BC ------> 6product
This equation suggests that the ratio of gas A to gas C is 3:2.
Based on the ratios obtained from the balanced equations, we can now calculate the actual relative masses of the elements.
Let's assume the relative mass of gas A (AB) is x, gas B (B) is y, and gas C (BC) is z.
From the ratios:
6x + y = 4
x + 4z = 4
3x + 2z = 6
Solving these equations simultaneously will give us the values of x, y, and z, which represent the actual relative masses of the elements in their simplest form.