Determine the empirical formulas of the compounds with the following compositions by mass: (a) 42.1% Na, 18.9% P, and 39.0% O

Video transcript

Hey everyone. Our question here wants us to determine the correct empirical formula for the following substance with 68.10% rubidium, 12.78% sulfur and 19.12% oxygen. A key to this question is recognized that we have 100% of our compound present. So we can go ahead and assume that we have 100 g. Starting off with rubidium, we have 68 0.10 g of rubidium And we want to convert this into moles by using its atomic mass of 85.47 graham's Permal. And when we calculate this out, We end up with a total of 0.7968 mol of rubidium. Moving on to Sulfur, we have 12.78 g of sulfur. And again, using our atomic mass of Sulfur, we know that we have 32.07 g per one mole. And when we calculate this out, We end up with a total of 0.3985 mol of Sulfur. Next looking at oxygen, we have 19.12g of oxygen And we can use oxygen's atomic mass, which is 16.0 g of oxygen per one mole of oxygen. And when we calculate this out, we end up with a total of 1.195 mole of oxygen. Now, in order to find our empirical formula, we need to divide our moles by the least amount of moles that we have here, and in this case it would be our mole of sulfur. So dividing each one x 0.3985, we're going to end up with a total of two of rubidium, one of sulfur And three of oxygen. So our empirical formula is going to be two of rubidium, one of sulfur and three of oxygen. So we end up with the following formula. So I hope this made sense and let us know if you have any questions.