Combustion of a hydrocarbon in a limited supply of air...

Combustion of a hydrocarbon in a limited supply of air produces CO in addition to CO2 and H2O. When 1.90 g of a particular hydrocarbon was burned in air, 1.83 g CO, 2.88 g CO2, and 2.95 g H2O were produced. What is the empirical formula of the hydrocarbon? empirical formula: Calculate the mass of oxygen used in the reaction. mass: Calculate the total mass of oxygen that would have been required for complete combustion. mass:

Transcript

00:01 In the first part of the question, we are assigned to the task to determine the empirical formula of the hydrocarbon.

00:14 It is given that the mass of hydrocarbon is 1 .90 gram.

00:20 We have 1 .83 gram of co, 2 .88 gram of co2 .85 gram of co2 and 2 .95 grams.

00:31 Gram of h2o.

00:35 Now we know that mols is equals to mars divided by molar mass, therefore moles of co is equal to 1 .83 divided by 28 .01, which is equal to 0 .0653 malls.

00:51 Similarly, the malls of co2 is equal to 2 .88 divided by 44, which is equal to 0 .0653.

01:02 The malls of h2o is equal to 2 .95 divided by 18 .02 and it is equal to 0 .1637 malls.

01:18 Now the malls of carbon is equal to 0 .1307 moles, that is the malls of carbon dioxide plus malls of carbon monoxide.

01:33 Moles of hydrogen is equal to 2 multiply 0 .1637, which is equal to 0 .3274 more.

01:44 Now, the ratio of carbon to hydrogen as the hydrocarbon contains carbon and hydrogen, 0 .1307 is to 0 .3274, that is 1 .304, that is 1 .0 to 2 .74, that is 1 .0 to 2 .74 .4.

02:01 In order to convert into a whole number multiply it by 2, we get 2 is to 5.

02:09 Therefore, it is c2h5.

02:12 The empirical formula of the hydrocarbon is c2h5.

02:17 In the second part, we have to calculate the mass of oxygen.

02:25 Now, the mass of oxygen in co is equal to 0 .0653.

02:31 The mass of oxygen in co2 is equal to 0 .1308.

02:38 The mass of oxygen in h2o is equal to 0 .1637, which is equals to 0 .3598.

02:52 This is the malls...