The percent composition by mass of copper and oxygen in copper(I) oxide is determined to be  % copper and  % oxygen. Round your answers to two decimal places.M Cu = 63.55 g mol-1M O = 16.00 g mol-1

It seems like you've missed providing the actual percentages of copper and oxygen in copper(I) oxide. However, I can explain how to calculate the percent composition by mass.

1. First, you need to know the molecular formula of copper(I) oxide, which is Cu2O. This means there are 2 atoms of copper (Cu) and 1 atom of oxygen (O) in one molecule of copper(I) oxide.

2. Next, calculate the molar mass of Cu2O. The molar mass of copper is given as 63.55 g/mol and there are 2 atoms of copper in the formula, so the total mass of copper is 2 * 63.55 g/mol = 127.1 g/mol. The molar mass of oxygen is given as 16.00 g/mol. So, the molar mass of Cu2O is 127.1 g/mol + 16.00 g/mol = 143.1 g/mol.

3. To find the percent composition by mass of copper in Cu2O, divide the total mass of copper by the molar mass of Cu2O and multiply by 100%. So, (127.1 g/mol ÷ 143.1 g/mol) * 100% = 88.82%.

4. Similarly, to find the percent composition by mass of oxygen in Cu2O, divide the mass of oxygen by the molar mass of Cu2O and multiply by 100%. So, (16.00 g/mol ÷ 143.1 g/mol) * 100% = 11.18%.

So, the percent composition by mass of copper and oxygen in copper(I) oxide is approximately 88.82% copper and 11.18% oxygen.