Compound X contains only carbon, hydrogen, nitrogen, and chlorine. When 1.00 g of X is dissolved in water and allowed to react with excess silver nitrate, AgNO3, all the chlorine in X reacts and 1.95 g of solid AgCl is formed. When 1.00 g of X undergoes complete combustion, 0.900 g of CO2 and 0.735 g of H2O are formed. What is the empirical formula of X?

Video transcript

Hey everyone, we're told that only carbon, hydrogen, nitrogen and chlorine are present in compound X. When excess silver nitrate is added to 1.52 g of sample X. After it has been dissolved in water, all of the chlorine and X reacts forming 3.1 g of solid silver chloride, Complete combustion of 1.52 g of sample x. results in the formation of 1.3 g of carbon dioxide and 0.95 g of water. Identify the empirical formula of sample X. First let's go ahead and determine the moles of each element. For sample X. Starting with our mole of chlorine, we were told that we had 3. g of silver chloride. Using the molar mass of silver chloride, we know that we have 143.32g of silver chloride Per one Mole of Silver Chloride. Next looking at our multiple ratios, we know that we have one mole of silver chloride and per one mole we have one mole of chlorine. When we calculate this out, we end up with a value of 0.02, one mole of chlorine. Now let's go ahead and determine the mass of chlorine. And the reason why we want to determine the mass of chlorine is because we will use it later to determine the mass of nitrogen. So again, starting with the value, we just calculated of 0.02, one Mole of chlorine. We're going to use chlorine molar mass. And looking at our periodic table, we see that one mole of chlorine contains 35.453g of chlorine. So when we calculate this out, we end up with a value of 0.74458 g of chlorine. Moving on to the mole of carbon and sample X, We were told that we had 1.39 g of carbon dioxide. Again, we're going to take the same steps as we did previously. And using carbon dioxides molar mass, we know that we have 44.01 g of carbon dioxide per one mole of carbon dioxide. Next, using our multiple ratios per one mole of carbon dioxide, we have one mole of carbon. So when we calculate this out, We end up with a value of 0.03158 mole of carbon. Next determining the mass of carbon. We're going to take 0.03 158 mole of carbon. And using Carbons Mueller mass, we know that we have 12.01 g of carbon Per one mole of carbon. And when we calculate this out, we end up with 0.379 g of carbon. Now let's go ahead and determine the mole of hydrogen in sample X. Starting with 0.95 g of water, We know that we have 18.016g of water Per one mole of water. Using our multiple ratios, we can see that one mole of water contains two mole of hydrogen. So when we calculate this out, We end up with 0.10546 mole of hydrogen. Now to calculate the mass of hydrogen, We're going to take 0.10546 mole of hydrogen. And using hydrogen molar mass, we know that we have 1.8 g of hydrogen per one mole of hydrogen. So when we calculate this out, we end up with a value of 0. g of hydrogen. Now let's go ahead and determine the mass of nitrogen. We can do so by taking 1. g Since we were told we had a complete combustion of 1.52 g of sample X. And we're going to subtract The mass of each element. So we have 0.74458 g of chlorine. And we also have 0.37935 g of carbon. And we have 0.1063 g of hydrogen. So when we calculate this out, we end up with 0.28977 g of nitrogen. Now using our grams of nitrogen to determine our mole of nitrogen, We're going to take 0.28977 g of nitrogen. And using nitrogen molar 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 0.02068 mole of nitrogen. Now let's go ahead and look at our mole ratios. So listing it all out here, We had 0.0-1 mole of chlorine And we also had 0.02068 mole of nitrogen. We had 0.10546 mole of hydrogen. And we had 0.03158 mole of carbon. Now comparing all of these values, we can see that our smallest mole is going to be our nitrogen. So we're going to divide each of these values by 0.02068. And this will get us to about one of chlorine, one of nitrogen, five of hydrogen and about 1.5 of carbon. So this gets us to a formula of C1. H five NCL. But since we can't have decimals for a subscript, We're going to multiply our formula by two. So each subscript will be multiplied by two in order to get a whole number. So our final formula is going to be C three, H 10 and two cl two. Now I hope this made sense. And let us know if you have any questions