An herbicide is found to contain only C, H, N, and Cl.
The compl...

Question

An herbicide is found to contain only C, H, N, and Cl. 
The complete combustion of a 100.0-mg sample of the 
herbicide in excess oxygen produces 83.16 mL of CO2 and 
73.30 mL of H2O vapor expressed at STP. A separate analysis 
shows that the sample also contains 16.44 mg of Cl. 
(b) Calculate its empirical formula.

Accepted Answer

Hey everyone in this example, we're told that a compound used as a preservative is made up of only carbon, hydrogen and oxygen and sodium. We're told that at standard temperature and pressure. A sample of the compound. Main 100 mg was burned with excess oxygen after the reaction is complete. 108, 108.88 ml of carbon dioxide and 38. ml of water were produced in a separate analysis. It was found that the sample contains 15.96 mg of sodium. We need to determine the molecular formula of this preservative. So based on the fact that the prompt gives us information on the atoms carbon dioxide and water because we know that this preservative that we need to find the molecular formula for is made up of the atoms carbon and hydrogen. We can go ahead and use these two compounds to determine the moles of each of these atoms from these two compounds. So beginning with moles of carbon from carbon dioxide, we're going to take that volume of carbon dioxide and we want to convert it to leaders. That is also due to the fact that to find our moles of our gas, we're going to reformulate our ideal gas equation so that we can find the molds of our gas equal to the pressure, multiplied by our volume and divided by r gas constant. R multiplied by temperature And we should recall that volume in this equation should be in leaders. So we're going to take the 108.88 ml and convert that to leaders by recalling that our prefix milli gives us 10 to the negative 3rd power leaders for one mil leader. So now we're able to cancel our units of middle leaders leaving us with leaders the proper unit for volume. And this is going to give us a volume of 0. 888 leaders of our C. 02. Now that we have this volume in the proper form, we can move into our calculation for moles of carbon. So we're going to find that our moles of carbon is equal to our pressure which we will assume because we're at standard pressure and temperature is going to be one A. T. M. We're going to multiply this by our volume which we just converted to leaders. So .10888 leaders. Then we're going to divide this by our gas constant R. Which we should recall is equal to 0. Leaders Times A T. M's divided by moles, times kelvin. And because we're at standard temperature we should recall that that's going to be 273.15 Kelvin. So what this is going to give us as a result of 4.8575 times 10 to the negative third power moles of our carbon atom from C. 02 and to show how the units canceled out. We would go ahead and cancel our A. T. M. S as well as leaders as well as kelvin's leaving us with moles, which is what we wanted for our final unit. So we're going to do the same thing and find our moles of carbon from or sorry, our moles of hydrogen rather since we already know most of carbon. So we want to find moles of hydrogen from the water given in the prompt. So we're given our volume for water in the prompt as 38.88 mL. We want to convert from milliliters to liters by recalling our prefix milli gives us 10 to the negative third power leaders. Now we're able to cancel out leaders leaving us with the proper volume or the proper unit for leaders for volume. Sorry, this gives us a value equal to 0.03888 l of our water molecule for the sorry, of the volume for a water molecule. And now we can find our moles of hydrogen by using our reorganized ideal gas equation so that we can take the pressure, which again is standard pressure being one A. T. M. Multiplied by our volume for our water molecule being 0.3888 liters that we calculated above. And dividing this by our gas constant R. Again, which you recall 0.8206 leaders times ATMs divided by moles times kelvin And then multiplying this by our standard temperature as to 73.15 Kelvin. So canceling out our units, we can get rid of kelvin, we can get rid of a T. M's leaders and we're left with moles. And what we're going to get here is a value equal to one seven 346 times 10 to the negative third power moles of our water molecule. So this gives us our molds of our water molecule and we want to again find molds of our hydrogen atom. We should recognize that our water molecule has to atoms of hydrogen in the molecule. So we would write out the ratio between hydrogen and water as a 1 to 2 ratio. And so we can take this moles of our water molecule 1. times 10 to the negative third power moles of water and multiply it by this ratio. So we know that we have in one mole of water, two moles of our hydrogen adam. And so multiplying out enumerators, we're going to get a value equal to 3.4692 times 10 to the negative third power moles of the hydrogen atom. We also want to make note of our mass of carbon from carbon dioxide as well as our mass of hydrogen from the water molecule. So for finding our mass of carbon From CO2, We're going to take our moles of carbon which above we found as 4. times 10 to the negative third power moles of carbon. And we're going to multiply this by from going, we're going to convert from moles of carbon to grams of carbon by recalling from the periodic table our molar mass for carbon For one mole of carbon, we would say that that's equal to 12.01 g of carbon. So we can cancel out our units of moles leaving us with g, which is what we want for mass. And we're going to convert two mg since in the prompt above were given information on milligrams for our masses of sodium as well as for our mass of oxygen. So we're going to convert from grams to milligrams by recalling that our prefix milli tells us that we have for one mg, 10 to the negative third power grams. So now we're able to cancel our units of grams, leaving us with milligrams as our final unit of mass. And this is going to give us a value equal to 58. mg of our carbon atom. Doing the same thing for our mass of hydrogen from water. We're going to take the moles of our or the moles of hydrogen from our water, which we found to be 3.4692 times 10 to the third power moles of hydrogen. We're going to multiply this by our molar mass of hydrogen, which you recall is in grams per mole. And from our periodic tables, we would see that for one mole of hydrogen, we have a mass of 1.08 g of hydrogen. So now we're left with canceling out moles and we're going to go from grams to milligrams again by recalling our prefix milli tells us that for one mg we have 10 to the negative third power grams. So now we're able to cancel out Graham's leaving us with milligrams. And this gives us a value equal to 3. mg of our hydrogen atom. So so far, the prompt tells us the mass of our sodium atom and milligrams. We just found our mass of carbon and hydrogen and milligrams. And now we need to find our mass of our oxygen atom in milligrams Because that was not given to us in the prompt. And so what we're going to do is because the prompt tells us that we have 100 mg of our unknown compound, which will just say unknown here. That is burned in excess with oxygen. We're going to subtract this from the sum of all of the masses of our carbon, hydrogen and sodium atoms. So for carbon, we said we have 58.33 mg, we're adding that to our mass of hydrogen as 3.50 mg. And then we'll be adding that from the prompt of our mass of sodium given as 15.96 mg of sodium. So this was given to us in the prompt and what we're going to get from these sums here as well as the difference from the total mass of our unknown compound is a difference equal to 22. mg of oxygen. And so now we want to go ahead and find moles of oxygen now that we know its mass. And so what we can say is that because we know its mass is 22.1 mg. We're going to convert from milligrams into grams by recalling again our prefix milli tells us that we have for one mg, 10 to the negative third power grams. Now we're able to cancel out milligrams and move from grams into moles by recalling our molar mass from the periodic table for an atom of oxygen, which tells us that we would have for one mole of of oxygen, 16 g of oxygen. And so now we're able to cancel out grams leaving us with most which is what we want and what we're going to get here is 1.3881 times 10 to the negative third power moles of oxygen. So we're going to follow the same process to find moles of our sodium atom in this compound Above. The prompt tells us that the mass of sodium is 15. mg of sodium. We're going to convert from milligrams to grams just like above. So for one mg we said we have 10 to the negative third power grams. We cancel out milligrams and lastly moved from grams of sodium two moles of sodium. And so we would recall from the periodic table, sodium, smaller mass equal to 23 g of sodium for one mole of sodium were able to cancel out grams. We're left with moles and what we're going to get is a value of 6.9391 times 10 to the negative fourth power moles of our sodium atom. And based on the smallest number of moles from one of the given atoms, we would conclude that our moles of sodium is the smallest number of moles. And so this is what we can use to determine our molecular formula. So from above, we stated that we have 4.8575 times 10 to the negative third power moles of carbon. And we're going to divide this by our smallest number of moles which came from our sodium atom as 6.9391 times 10 to the negative fourth power moles of sodium Dividing this out. We're going to get a value for this quotient equal to a value of 7.00. Which we can say is just equal to about seven. So this would be our subscript of the carbon atom in our molecular compound. So moving on to our moles of hydrogen, which above, we found to be equal to 3.4692 times 10 to the negative third power moles of hydrogen. We're going to divide this by the smallest number of moles 6.9391 times 10 to the negative fourth power moles of sodium. And this gives us a value equal to about size. Moving on, we have our moles of oxygen, which above we stated is 1.3881 times 10 to the negative third power moles of oxygen, dividing this out by the smallest number of moles 6.9391 times 10 to the negative fourth power moles of sodium. We're going to get a value equal to about two. And then lastly, we have our moles of our sodium atom which again we agreed is 6.9391 times 10 to the negative fourth power moles of sodium divided by the same value, which happens to be our smallest number of moles. And this gives us one. And so what we can say is that therefore our molecular formula is going to be C seven H five and then N. A. At the end for our preservative compound. So, this will be our final answer for our molecular formula. I hope that everything I explained was clear. If you have any questions, leave them down below and I will see everyone in the next practice video

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