What is the empirical formula for a compound which contains 67.1 percent zinc and the rest is oxygen
Take a 100 g sample which means
67.1 g Zn
100-67.1 = 32.9 g oxygen.
moles Zn = 67.1/atomic mass Zn = ?.
moles O = 32.9/16 = ?
Now find the ratio of the two elements to each other in small whole numbers with the smallest number no less than 1.00. The easy way to do that is to divide the smaller number of moles by itself thereby assuring you of obtaining 1.00 for that one. Then divide the other number by that same small number. Then round to whole numbers.
1000
ZnO2
To determine the empirical formula of a compound, you need to know the mass or percent composition of each element in the compound.
In this case, the compound contains 67.1% zinc and the remaining percentage is oxygen. To determine the empirical formula, we can assume that we have 100 grams of the compound. This allows us to convert the percentages into grams.
Since the compound is 67.1% zinc, we have 67.1 grams of zinc. The remaining 32.9 grams (100 - 67.1) is oxygen.
The next step is to convert the grams of each element into moles. We can do this by dividing the mass by the molar mass of the respective element.
The molar mass of zinc (Zn) is 65.38 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol.
For zinc:
67.1 g Zn × (1 mol Zn / 65.38 g Zn) ≈ 1.026 mol Zn
For oxygen:
32.9 g O × (1 mol O / 16.00 g O) ≈ 2.056 mol O
The final step is to find the simplest whole number ratio of the elements. By dividing both values by the smaller value, we can find the ratio:
1.026 mol Zn ÷ 1.026 ≈ 1 mol Zn
2.056 mol O ÷ 1.026 ≈ 2 mol O
So, the empirical formula for the compound is ZnO, indicating that it contains one zinc atom and two oxygen atoms.