A compound composed of carbon, hydrogen, and chlorine contains...

Question

A compound composed of carbon, hydrogen, and chlorine contains 4.19 x 10²³ hydrogen atoms. If 9.00 g of the compound also contains 55.0% chlorine by mass, what is the empirical formula?

Accepted Answer

Here the practice question states: A compound composed of carbon, hydrogen, and chlorine contains 4.19 times 10 to the 23 hydrogen atoms. If 9 grams of the compound also contains 55% chlorine by mass, what is the empirical formula? Now, sometimes you'll run into a question and based on its wording may seem very confusing. But remember, just first look to see what are they asking me to find. They want me to find the empirical formula. Then rely on the steps that you've learned to help calculate the empirical formula. Remember, Step 2, we need to find the masses in grams of each of the elements. So let's do that first. Here we have hydrogen atoms. So what we need to do here is we need to convert those hydrogen atoms into grams. So we have 4.19 times 10 to the 23 hydrogen atoms. Atoms means we're talking about a single atomic element so it's just H by itself. What we're going to do next is we're going to go from atoms to moles. Remember that means that we have to use Avogadro's number. So 1 mole of hydrogen is 6.022 times 10 to the 23 hydrogen atoms. And remember, I've said this in earlier videos. When it comes to working things out like this, if you have something written in scientific notation, it's best to put it in parenthesis in your calculator because based on the model that you're using, you may get an entirely wrong answer. So, again, if numbers are written in scientific notation, it's scientific notation, so a number times 10 to any power, put it in parenthesis. So now, add a number times 10 to any power, put it in parenthesis. So now, atoms cancel out. We have moles of hydrogen and we're going to say that for every 1 mole of hydrogen, it's 1.008 grams hydrogen. So moles of h cancel out and we have our grams as 0.7013 grams hydrogen. Alright. So we have hydrogen figured out. Remember this compound is composed of carbon and chlorine as well. They're telling us that our sample weighs, or compound weighs 9 grams and 55% of that total mass is chlorine. So we'd say that we have 9 grams total. We cannot use the percentage form when it comes to math, we have to use the decimal form. So you divide this by a 100 and that gives me 0.5 5. So it's 0.55 times 9, so that gives me 4.95 grams of chlorine. So now we've found the grams of hydrogen, the grams of chlorine. How do we figure out the grams of carbon? Well, remember we said that our compound weighs 9 grams. So those 9 grams are of the carbon, hydrogen, and chlorine altogether. We just found out the masses of hydrogen and chlorine, so subtract those from 9 and the difference which you have left will have to represent the grams So we So we have left at the end is 3.3487 grams of carbon. So this originally difficult looking question is pretty approachable now because we have the grams of each of the elements. So all we're going to do is follow the steps that we know. We're going to change those grams into moles, divide them all by the smallest moles and hopefully be able to get our empirical formula at the end. Alright. So, we're gonna start out with carbon. So we have those grams of carbon. Oops. So we have 3.3487 grams carbon. We have point 7013 grams hydrogen. And we have 4.95 grams chlorine. I'm writing in this order because that's the order, they're given to us within the question. Carbon, hydrogen and then chlorine. Alright. So, 1 mole of carbon weighs 12.01 grams carbon. One mole of h is 1.008 grams h. And then, 1 mole of chlorine is 35.45 grams chlorine. All the grams will cancel out, and we'll have moles at the end. Remember, at this point we need to make sure that our moles had at least 4 decimal places carbon. Hydrogen comes out to be 0.6958 moles h. And then chlorine comes out to be 0.1396 moles of chlorine. Next we divide all the moles that we've gotten by the smallest mole answer. Which would be the 0.1396. So that would give us 1 chlorine. And then when we divide hydrogen by that number, we get 4.984. And because it's 0.9, 9, we can round up to 5, so that's 5 h. And then carbon, when we divide it by that, we get 1.9 97 carbon. Again, it's 0.9 so I can round up to 2. So we have 2 carbons, 5 hydrogens and one chlorine. That means that our empirical formula here would be c2h5cl. So again, there will be times when you're gonna run into a question that seems very confusing by the way it's written. 1st, find out what exactly are they asking you to find and then rely on the steps that you've learned to help you get to the answer. Here, all we have to do is convert all the information we have to the grams of each of the elements. They only gave us information on carb on chlorine and hydrogen. We found their grams. We know that the complete mass of the compound was 9 grams so we knew that, the grams of chlorine and hydrogen if we subtract it from the 9 grams, that would give us the mass of the missing carbon. Once you have the grams of everything, we follow the basic steps that we've learned to find our empirical formula. So keep this in mind and remember the best way to approach any question that seems confusing is to first find out what they're asking you to do. And then if you know any steps, follow those steps as closely as you can to get to your final answer.

Introduction to Chemistry 때문에 힘들어하고 계신가요?

A compound composed of carbon, hydrogen, and chlorine contains 4.19 x 10²³ hydrogen atoms. If 9.00 g of the compound also contains 55.0% chlorine by mass, what is the empirical formula?

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Empirical Formula 연습 세트

Introduction to Chemistry 때문에 힘들어하고 계신가요?

A compound composed of carbon, hydrogen, and chlorine contains 4.19 x 1023 hydrogen atoms. If 9.00 g of the compound also contains 55.0% chlorine by mass, what is the empirical formula?

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Empirical Formula 연습 세트

A compound composed of carbon, hydrogen, and chlorine contains 4.19 x 10²³ hydrogen atoms. If 9.00 g of the compound also contains 55.0% chlorine by mass, what is the empirical formula?