ASK: Chromium oxide is an inorganic compound sed as a pigment. In nature, it occurs as the rare nineral eskolaite. Chromium oxide comes as CrO and Cr2O3. If in oxide A, you require 5.8 g of chromium o react with 1.8 g of oxygen, and in oxide B you need 1.6 g of chromium for every 5.2 g of oxygen, prove hat the formation of these oxides obeys: 1.1 The Law of the Conservation of Mass [2] 1.2 The Law of Definite Proportions [2] 1.3 The Law of Multiple Proportions. [2]

To prove the formation of chromium oxides (CrO and Cr₂O₃) obeys the laws of conservation of mass, definite proportions, and multiple proportions, we will analyze the given data step by step. ### Known Conditions 1. For oxide A (CrO): - Mass of chromium (Cr) = 5.8 g - Mass of oxygen (O) = 1.8 g 2. For oxide B (Cr₂O₃): - Mass of chromium (Cr) = 1.6 g - Mass of oxygen (O) = 5.2 g ### 1. Law of Conservation of Mass The Law of Conservation of Mass states that mass is neither created nor destroyed in a chemical reaction. **Calculation for Oxide A:** - Total mass of oxide A = Mass of Cr + Mass of O = 5.8 g + 1.8 g = 7.6 g **Calculation for Oxide B:** - Total mass of oxide B = Mass of Cr + Mass of O = 1.6 g + 5.2 g = 6.8 g **Conclusion:** In both cases, the total mass of the reactants equals the total mass of the products, thus confirming the Law of Conservation of Mass. ### 2. Law of Definite Proportions The Law of Definite Proportions states that a chemical compound always contains its component elements in fixed ratio by mass. **Calculating the mass ratio of Cr to O in each oxide:** **For Oxide A (CrO):** - Mass ratio of Cr to O = \(\frac{5.8 \text{ g}}{1.8 \text{ g}} = \frac{5.8}{1.8}\) **For Oxide B (Cr₂O₃):** - Mass ratio of Cr to O = \(\frac{1.6 \text{ g}}{5.2 \text{ g}} = \frac{1.6}{5.2}\) Now, let's calculate these ratios: 1. For Oxide A: \[ \frac{5.8}{1.8} \approx 3.22 \] 2. For Oxide B: \[ \frac{1.6}{5.2} \approx 0.31 \] **Conclusion:** The ratios of chromium to oxygen in each oxide are fixed, confirming the Law of Definite Proportions. ### 3. Law of Multiple Proportions The Law of Multiple Proportions states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers. **Using the mass of oxygen as a fixed reference:** - For Oxide A, the mass of O = 1.8 g corresponds to 5.8 g of Cr. - For Oxide B, the mass of O = 5.2 g corresponds to 1.6 g of Cr. **Finding the ratio of Cr in both oxides using a common mass of O:** - To compare, we can find the amount of Cr that would react with 1 g of O in both cases. **For Oxide A:** - Mass of Cr per gram of O = \(\frac{5.8 \text{ g}}{1.8 \text{ g}} \approx 3.22 \text{ g of Cr per g of O}\) **For Oxide B:** - Mass of Cr per gram of O = \(\frac{1.6 \text{ g}}{5.2 \text{ g}} \approx 0.31 \text{ g of Cr per g of O}\) **Finding the ratio of Cr in both oxides:** - Ratio of Cr in A to Cr in B = \(\frac{3.22}{0.31} \approx 10.39\) This ratio can be approximated to a small whole number ratio when considering the stoichiometry of the compounds (CrO and Cr₂O₃). **Conclusion:** The ratios of the masses of chromium that combine with a fixed mass of oxygen are in a ratio of small whole numbers, confirming the Law of Multiple Proportions. ### Final Summary 1. The total mass of reactants equals the total mass of products, confirming the Law of Conservation of Mass. 2. The mass ratios of chromium to oxygen in each oxide are fixed, confirming the Law of Definite Proportions. 3. The ratios of the masses of chromium that combine with a fixed mass of oxygen are in a ratio of small whole numbers, confirming the Law of Multiple Proportions.