Empirical formula 88.8% copper 11.2%oxygen

empirical-formula

#1

empirical formula 88.8% copper 11.2%oxygen

Answer:

OK, I’m assuming that those percentages are measured by mass. Here’s one relatively simple way of doing it. Since no actual mass is given, but that ratio would remain constant, you can assume that the total mass is 100g. In that case you would have 88.8g of Copper and 11.2g of Oxygen per 100g total. So, you find out how many moles of each this would be. 88.8g of Copper is about 1.4moles, and 11.2g of Oxygen is about .7moles. Do you notice the relationship? There are twice as many moles of Copper as there are moles of Oxygen. So the empirical formula is Cu2O.

The important concept in this question is knowing the difference between empirical and molecular formulas, and the significance of percent by mass. Empirical formulas are simply the ratio by moles. However if you had used the percent by mass you would’ve gotten something on the order of Cu8O, which is not very correct. A molecular formula on the other hand wants the literal ratio per mole, so you would need the molar mass to solve.