last one, i PROMISE
A compound containing only C, H, O and Cr was subjected to combustion analysis. A sample of 10.88 mg produced 22.43 mg of CO2 and 4.59 mg of H2O. The chromium in the sample was precipitated as 14.09 mg of Ag2CrO4. Determine the empirical formula of the compound and match each element with its correct subscript from the drop box.
%CO2 = (23.43/10.88)*100.
%H2O = (4.59/10.88)*100.
For Cr, convert 14.09 mg Ag2CrO4 to g, then to mols Ag2CrO4, then to mols Cr, then to %Cr using the same format as above.
%O = 100% - %CO2 - %H2O - %Cr.
Now take a 100 g sample which will give you xx g CO2, yy g H2O, zz O and ww g Cr.
Convert g to mols C, H, O, and Cr.
Then find the ratio of each. The easy way to do this is to divide the smallest number by itself, then divide all the other mols by that number. Post your work if you need more help.
This should get you started.
Determine
I hope you didn't spend much time on this for I am way off base, including some typos. You need to look up the actual molar masses. I have estimated some of them. I just don't know what I was thinking. (I wasn't thinking.)
%C =[22.42 x (1 mol C/1 mol CO2)/10.88]*100 = [22.42 x (12/44)/10.88]*100 = xx
%H = [4.59 x (2 mol H/1 mol H2O/10.88]*100 = [4.59 x (2/18)/10.88] = yy
%Cr = [14.08 x (1 mol Cr/1 mol Ag2CrO4)/10.88]*100 = [14.08 x (52/331.7)/10.88]*100 = zz
%O = %C - %H - %Cr = ww
Now take a 100 g sample which will give you xx g C, yy g H, zz g Cr and ww g O
Then find the ratio of each. The easy way to do this is to divide the smallest number by itself, then divide all the other mols by that number. Post your work if you need more help.
After you take the 100 g sample, you want to take the grams of each and convert to mols, then find the ratio of the mols.
So,
xx/12= mols C
yy/1 = mols H
zz/52 = mols Cr
ww/16 = mols O
Now go back and read about dividing the smallest number etc.
To determine the empirical formula of the compound, we need to calculate the ratios of the elements present in the compound. We are given the masses of carbon dioxide (CO2), water (H2O), and silver chromate (Ag2CrO4) produced during the combustion analysis.
1. Let's start by finding the moles of each element:
a. Moles of carbon (C):
- Molar mass of CO2 = 12.01 g/mol (atomic mass of carbon) + 2 * 16.00 g/mol (atomic mass of oxygen) = 44.01 g/mol
- Moles of CO2 = 22.43 mg / 44.01 g/mol = 0.5092 mol
b. Moles of hydrogen (H):
- Molar mass of H2O = 2 * 1.01 g/mol (atomic mass of hydrogen) + 16.00 g/mol (atomic mass of oxygen) = 18.02 g/mol
- Moles of H2O = 4.59 mg / 18.02 g/mol = 0.2545 mol
c. Moles of chromium (Cr) in Ag2CrO4:
- Molar mass of Ag2CrO4 = 2 * 107.87 g/mol (atomic mass of silver) + 52.00 g/mol (atomic mass of chromium) + 4 * 16.00 g/mol (atomic mass of oxygen) = 331.84 g/mol
- Moles of Ag2CrO4 = 14.09 mg / 331.84 g/mol = 0.0425 mol
2. Next, let's find the moles of oxygen (O) present. Since the oxygen in both CO2 and H2O come from the same source, we add their moles together:
- Moles of O = 2 * moles of CO2 + moles of H2O = 2 * 0.5092 mol + 0.2545 mol = 1.2729 mol
3. Finally, we can determine the empirical formula by converting all the moles to the simplest whole-number ratios.
a. Carbon (C): There are approximately 0.5092 mol of carbon.
b. Hydrogen (H): There are approximately 0.2545 mol of hydrogen.
c. Oxygen (O): There are approximately 1.2729 mol of oxygen.
d. Chromium (Cr): There are approximately 0.0425 mol of chromium.
To find the simplest whole-number ratio, we divide all the values by the smallest number of moles, which in this case is 0.0425 mol (the chromium amount):
a. Carbon (C): 0.5092 mol / 0.0425 mol = 12
b. Hydrogen (H): 0.2545 mol / 0.0425 mol = 6
c. Oxygen (O): 1.2729 mol / 0.0425 mol = 30
d. Chromium (Cr): 0.0425 mol / 0.0425 mol = 1
Therefore, the empirical formula of the compound is C12H6O30Cr.