Understanding the Problem
We have two metal oxides. The first oxide has the formula M₃O₄, where M represents the metal. We know the percentage of oxygen in both oxides. We need to find the formula of the second oxide.
Steps
1. Determine the Metal's Molar Mass in the First Oxide:
* Let the molar mass of the metal (M) be 'x'.
* The molar mass of M₃O₄ = 3x + (4 * 16) = 3x + 64
* Percentage of oxygen in M₃O₄ = (64 / (3x + 64)) * 100 = 27.6
* Solve for 'x': 6400 = 82.8x + 1769.6
* 4630.4 = 82.8x
* x ≈ 55.8
2. Determine the Empirical Formula of the Second Oxide:
* The second oxide contains 30% oxygen by mass.
* This means if we assume a 100g sample, there are 30g of oxygen and 70g of the metal.
* Convert grams to moles:
* Moles of oxygen = 30g / 16g/mol = 1.875 mol
* Moles of metal = 70g / 55.8g/mol ≈ 1.25 mol
* Divide the moles of each element by the smallest number of moles to find the simplest whole number ratio:
* O: 1.875 / 1.25 = 1.5
* M: 1.25 / 1.25 = 1
* Since we can't have fractional subscripts, multiply both by 2 to get whole numbers:
* O: 3
* M: 2
The Formula of the Second Oxide
The formula of the second oxide is M₂O₃.
