Two oxides of a metal contain 50% and 40% metal (M) respectively. If the formula of first oxide is MO2 the formula of second oxide will be MO2 MO3 M2O M2O5

The problem is about determining the formula of the second oxide based on the percentage of metal in two different oxides of the same metal.

Step 1: Let's assume the atomic weight of the metal M is x.

Step 2: In the first oxide (MO2), the percentage of metal is 50%. This means that the weight of the metal is half of the total weight of the compound. Since the formula of the compound is MO2, the total weight of the compound is x (weight of M) + 32 (weight of O2). So, we have the equation: x/(x+32) = 50/100.

Step 3: Solving the equation for x, we get x = 32. So, the atomic weight of the metal M is 32.

Step 4: In the second oxide, the percentage of metal is 40%. This means that the weight of the metal is 40% of the total weight of the compound. Let's assume the formula of the second compound is MxOy. So, the total weight of the compound is 32x (weight of M) + 16y (weight of O). So, we have the equation: 32x/(32x+16y) = 40/100.

Step 5: Simplifying the equation, we get x/(x+y) = 2/5.

Step 6: The only way to satisfy this equation is when x = 2 and y = 3. So, the formula of the second oxide is M2O3.