Question

Gravimetric Analysis -- Binary Mixture

Amounts of tin may be alloyed with copper to produce bronzes of varying composition and properties.
4.3330 g of a particular bronze, compounded of only Cu and Sn, can be quantitatively converted into a 5.4334-g mixture of the oxides CuO and SnO₂.
(Use the atomic masses: Cu 63.546, Sn 118.71, O 15.9994.)

What was the percentage (by mass) of tin in the alloy?

Answer 1

Ans. # Step 1: Let the mass of Cu and Sn in the bronze sample be X and Y grams, respectively.

            So, X g + Y g = 4.3330 g

            Or, X + Y = 4.3330               - equation 1

# Step 2:

I. Moles of Cu = Mass / MW = X g / 63.546 g mol^-1 = 0.0157366 X mol

1 mol Cu forms 1 mol CuO. So, moles of CuO formed = 0.0157366 X mol

Now,

            Mass of CuO formed = Moles x MW = 0.0157366 X mol x 79.5454 g mol^-1

                                                = 1.2517767 X g

II. Moles of Sn = Mass / MW = Y g / 118.71 g mol^-1 = 0.0084239 Y mol

1 mol Sn forms 1 mol SnO₂. So, moles of SnO₂ formed = 0.0084239 Y mol

Now,

            Mass of SnO₂ formed = Moles x MW = 0.0084239 Y mol x 150.7088 g mol^-1

                                                = 1.2695544 Y g

            III. Given, total mass of oxide formed = 5.4334 g

So,

            1.2517767 X g + 1.2695544 Y g = 5.4334 g

            Or, 1.2517767 X + 1.2695544 Y = 5.4334                     - equation 2

# Step 3: Comparing (equation 1 x 1.2517767) – equation 2-

            1.2517767 X + 1.2517767 Y = 5.4239484411

-1.2517767 X - 1.2695544 Y = -5.4334

---------------------------------------------------------------

                        -0.0177777 Y = -0.0094515589

                        Or, Y = 0.0094515589 / 0.0177777 = 0.53165

Therefore, mass of Sn in the bronze sample = 0.53165 g

# Step 4: % Sn (tin) in sample = (Mass of Sn / Mass of sample) x 100

                                    = (0.53165 g / 4.3330 g) x 100

                                    = 12.27 %

Answer 2

To find the percentage of tin in the alloy, we need to determine the masses of copper (Cu) and tin (Sn) in the original bronze sample.

Let's assume the mass of copper in the bronze is x grams. Then the mass of tin in the bronze would be (4.3330 g - x grams).

According to the given information, the 4.3330 g of the bronze is quantitatively converted into a 5.4334 g mixture of CuO and SnO2.

To calculate the masses of copper oxide (CuO) and tin dioxide (SnO2) in the mixture, we need to use their molar masses.

Molar mass of CuO = atomic mass of Cu + 1 * atomic mass of O = 63.546 g/mol + 15.9994 g/mol = 79.5454 g/mol

Molar mass of SnO2 = atomic mass of Sn + 2 * atomic mass of O = 118.71 g/mol + 2 * 15.9994 g/mol = 159.7098 g/mol

Let's assume the mass of CuO in the mixture is y grams. Then the mass of SnO2 in the mixture would be (5.4334 g - y grams).

Now we can set up an equation based on the conservation of mass:

Mass of copper + Mass of tin = Mass of CuO + Mass of SnO2

x + (4.3330 g - x) = y + (5.4334 g - y)

Simplifying the equation:

4.3330 g = 1.1004 g + x - y

Now we need to convert the masses of CuO and SnO2 to moles:

moles of CuO = mass of CuO / molar mass of CuO = y / 79.5454 g/mol

moles of SnO2 = mass of SnO2 / molar mass of SnO2 = (5.4334 g - y) / 159.7098 g/mol

According to the balanced equation of the reaction, the ratio between moles of copper and moles of tin is 1:1.

So, moles of Cu = moles of Sn.

Therefore, we can write:

y / 79.5454 g/mol = (4.3330 g - x) / 118.71 g/mol

Cross-multiplying and simplifying:

y * 118.71 g/mol = (4.3330 g - x) * 79.5454 g/mol

Now we can substitute the equation above into the previous equation:

4.3330 g = 1.1004 g + x - (4.3330 g - x) * 79.5454 / 118.71

Simplifying and solving for x:

4.3330 g = 1.1004 g + x - (4.3330 g - x) * 0.66996

4.3330 g = 1.1004 g + x - 2.8982 g + 0.66996 x

4.3330 g = -1.7978 g + 1.66996 x

6.1308 g = 1.66996 x

x = 6.1308 g / 1.66996 ≈ 3.6759 g (mass of copper in the bronze)

Now we can calculate the mass of tin in the bronze:

Mass of tin = 4.3330 g - Mass of copper = 4.3330 g - 3.6759 g