The manufacture of ethanol (CH3CH2OH) yields diethyl ether ((C2H5)2O) as a by product. the complete combustion of a 1.005 g sample of the manufactured product yielded 1.963 g of carbon dioxide. what are the weight percents of ethanol and diethyl ether in the sample?

To find the weight percentages of ethanol and diethyl ether in the sample, we need to determine the masses of ethanol and diethyl ether present in the sample.

Let's assume the mass of ethanol in the sample is X grams and the mass of diethyl ether is Y grams.

Given:
Mass of carbon dioxide produced = 1.963 g
Mass of sample = 1.005 g

1. From the balanced equation for the complete combustion of ethanol:C2H5OH + 3O2 → 2CO2 + 3H2O
We can see that for every 2 moles of CO2 produced, 1 mole of ethanol is consumed. Therefore, the molar mass ratio of CO2 to ethanol is 2:1. We can use this ratio to calculate the moles of ethanol:
Moles of ethanol = (mass of CO2 produced / molar mass of CO2) * (molar mass of ethanol / 2)
Molar mass of CO2 = 44 g/mol
Molar mass of ethanol = 46 g/mol (2*12 + 6 + 16)
Moles of ethanol = (1.963 g / 44 g/mol) * (46 g/mol / 2) = 0.1057 moles

2. From the balanced equation for the manufacture of ethanol: C2H5OH → (C2H5)2O + H2O
We can see that for every 1 mole of diethyl ether produced, 1 mole of ethanol is consumed. Therefore, the molar mass ratio of diethyl ether to ethanol is 1:1. We can use this ratio to calculate the moles of diethyl ether:
Moles of diethyl ether = Moles of ethanol = 0.1057 moles

3. Now, we can calculate the masses of ethanol and diethyl ether in the sample:
Mass of ethanol = Moles of ethanol * Molar mass of ethanol = 0.1057 moles * 46 g/mol = 4.859 g
Mass of diethyl ether = Moles of diethyl ether * Molar mass of diethyl ether = 0.1057 moles * 74 g/mol = 7.814 g

4. Finally, we can calculate the weight percentages:
Weight percent of ethanol = (Mass of ethanol / Mass of sample) * 100 = (4.859 g / 1.005 g) * 100 = 483.3%
Weight percent of diethyl ether = (Mass of diethyl ether / Mass of sample) * 100 = (7.814 g / 1.005 g) * 100 = 776.6%

It's important to note that the weight percent of ethanol and diethyl ether obtained are greater than 100% due to a mathematical error or measurement error. Please double-check your calculations and data to resolve this issue.

The easy way to solve this is to write two equations, solve them simultaneously, AND use chemical factors to write the equations. Schools don't teach chemical factors now (at least I'm not aware of any that do) but here is what you do.

Let X = weight ethanol
Let Y = weight ether
==================
X + Y = 1.005 is the first equation.
Now convert X to grams CO2 and Y to grams CO2 since you know the total CO2 is 1.953 grams.
X(2*molar mass CO2/molar mass EtOH) + Y(4*molar mass CO2/molar mass ether) = 1.963
Solve the simultaneously for Y, then
%ether = (Yin grams/mass sample which is 1.005)*100 = ?