An organic compound was found to contain only C, H, and Cl. When a 1.50-g sample of the compound was completely combusted in air, 3.52 g of CO2 was formed. In a separate experiment, the chlorine in a 1.00-g sample of the compound was converted to 1.27 g of AgCl. Determine the empirical formula of the compound.

Video transcript

Hey everyone, we're told that in an experiment 2.35 g of sample was combusted and produced 1.10 g of nitrogen dioxide. In another experiment 2.67 g of the same sample was converted to five g of lead to bromide. What is the empirical formula of the sample? If it is only composed of nitrogen, hydrogen and bromine first, we need to determine the percentages of nitrogen, hydrogen and bromine. Starting off with our percent bro. Ming, We can take the 5g of lead to bromide And we can use dimensional analysis to determine how many grams of roaming we have. So looking at led to bromides molar mass, we know that we have 367.01 g of lead to bromide per one mole of lead to bromide. Next looking at our multiple ratios, we can see that per one mole of lead to bromide. We have two mol of roaming. And lastly looking at bro means Mueller mass, we know that we have 79.904 g of bro. Mean Per one mole of grooming. So when we calculate this out and cancel out all of our units, We end up with 2.1772 g of roaming Not to convert this into g. We're going to take our grams of roaming which was 2.1772 g. And we're going to divide this by our sample. So it would be divided by 2.67 g of our sample. And we're also going to multiply this by 100% to get that percentage. So this comes up to 81.54, roaming. Now, let's go ahead and determine our percent of nitrogen. Again, we're going to take the same steps, but instead we're going to use 1.10 g of nitrogen dioxide Using nitrogen dioxide solar mass. We know that we have 46.01 g of nitrogen dioxide per one mole of nitrogen dioxide. Looking at our multiple ratios, we know that per one mole of nitrogen dioxide We have one mole of nitrogen And looking at nitrogen Mueller mass, we know that we have 14.01 g of nitrogen Per one mole of nitrogen. So when we calculate this out, we end up with a total of 0.3349 g of nitrogen. Now to get our percent nitrogen, we're going to take 0.3349 g of nitrogen And divided by 2.35 g of our sample. Again, we're going to multiply this by 100%. And this will get us to our percentage of 14.253% nitrogen. Now to calculate our percent hydrogen, we're going to do so through subtraction. So we have 100% and we're going to subtract 81.542% from roaming. And we're also going to subtract 14.25, from nitrogen. This will get us to 4.205% of hydrogen. Now to determine our empirical formula, we can go ahead and assume that we have 100 g of our sample. So converting those percentages into grams, we have 81.542 g of roaming. And using bro means Mueller mass. We know that we have 79.904 g of roaming Per one Mole of Browning. This will get us to 1.0205 mol of roaming. Next looking at Hydrogen, we had 4.205g of hydrogen And we know that we have 1.01 g of hydrogen per one mole of hydrogen. This will get us to a value of 4.1716 mole of hydrogen. And lastly looking at our nitrogen, we know that we have 14.253 g of nitrogen. And looking at our periodic table, we know that we have 14 g of nitrogen Per one mole of nitrogen. This will get us to 1.0181 mole of nitrogen. Now to get our mole ratio, we want to divide each of these values by the lowest value we have here. In this case it will be 1.018, 1 Mole. This will get us to one of nitrogen four of hydrogen and one of roaming. So our empirical formula comes up to NH four B r. Now, I hope this made sense. And let us know if you have any questions.