a 2.52-g sample of a compound containing only carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess O to yield 4.23 g of CO2 and 1.01 g of H2O as the only carbon and hydrogen containing products respectively.
Another sample of the same compound, of mass 4.14 g, yielded 2.11 g of SO3 as the only sulfur containing product.
A third sample, of mass 5.66 g, was burned under different conditions to yield 2.27 g of HNO3 as the only nitrogen containing product.
PLEASE HELP! THIS IS DUE VERY SOON AND I HAVE BEEN WORKING ON IT FOR HOURS BUT KEEP GETTING STUCK:(
What is the question? The empirical formula?
Convert 4.23g CO2 to g C and convert that to %C. Here is how you do that.
4.23g CO2 x (atomic mass C/molar mass CO2) = 4.23 X 12/44 = about 1.15g C, then %C = 1.15/mass sample)*100 = about 45.8%C
For g H it is 1.01g H2O x (2*1/18) = 0.112g and %H = (0.112/2.52)*100 = about 4.45%
You should redo these to confirm and/or obtain with better accuracy.
Do th same kind of thing for %S (in the different sample but % is %) and the same for %N. That leaves %O which is obtained by adding %C, %H, %N, %S and subtracting from 100.
Now take a 100 g sample which gives you the same numbers as the % but without the % sign.
Convert g each to mols by
mol = g/atomic mass and I assume you can take it from here. Post your work if you get stuck.I didn't work it through to get an answer but this should be straight forward from here.
To determine the empirical formula of the compound, we need to calculate the molar ratios of each element present. Here's how you can approach this problem step by step:
Step 1: Calculate the moles of carbon and hydrogen in the first sample.
- The molar mass of CO2 is 44.01 g/mol. Thus, the moles of carbon in the first sample can be calculated as:
Moles of carbon = Mass of CO2 / Molar mass of CO2
= 4.23 g / 44.01 g/mol
- The molar mass of H2O is 18.02 g/mol. Thus, the moles of hydrogen in the first sample can be calculated as:
Moles of hydrogen = Mass of H2O / Molar mass of H2O
= 1.01 g / 18.02 g/mol
Step 2: Calculate the moles of sulfur in the second sample.
- The molar mass of SO3 is 80.06 g/mol. Thus, the moles of sulfur in the second sample can be calculated as:
Moles of sulfur = Mass of SO3 / Molar mass of SO3
= 2.11 g / 80.06 g/mol
Step 3: Calculate the moles of nitrogen in the third sample.
- The molar mass of HNO3 is 63.01 g/mol. Thus, the moles of nitrogen in the third sample can be calculated as:
Moles of nitrogen = Mass of HNO3 / Molar mass of HNO3
= 2.27 g / 63.01 g/mol
Step 4: Calculate the moles of oxygen in each sample.
- To obtain the moles of oxygen, subtract the sum of moles of other elements (carbon, hydrogen, sulfur, and nitrogen) from the total moles present in the sample.
Step 5: Determine the smallest mole value among carbon, hydrogen, sulfur, nitrogen, and oxygen. This will be used as the reference value for calculating the whole number ratio.
Step 6: Divide the moles of each element by the smallest mole value to get the whole number ratio.
Step 7: Convert the whole number ratio to the empirical formula by using the subscripts of the elements.
That's the step-by-step process to determine the empirical formula of the compound based on the given data.
To solve this problem, we need to determine the empirical formula of the compound based on the given information.
Step 1: Calculate the moles of CO2 and H2O produced from the combustion of the first sample.
First, we need to convert the mass of CO2 and H2O to moles:
Molar mass of CO2 = 12.01 g/mol (carbon) + 2 * 16.00 g/mol (oxygen) = 44.01 g/mol
Molar mass of H2O = 2 * 1.01 g/mol (hydrogen) + 16.00 g/mol (oxygen) = 18.02 g/mol
Moles of CO2 = 4.23 g CO2 / 44.01 g/mol = 0.096 moles
Moles of H2O = 1.01 g H2O / 18.02 g/mol = 0.056 moles
Step 2: Calculate the moles of S and SO3 produced from the second sample.
Molar mass of SO3 = 32.07 g/mol (sulfur) + 3 * 16.00 g/mol (oxygen) = 80.06 g/mol
Moles of SO3 = 2.11 g SO3 / 80.06 g/mol = 0.026 moles
Step 3: Calculate the moles of N and HNO3 produced from the third sample.
Molar mass of HNO3 = 1.01 g/mol (hydrogen) + 14.01 g/mol (nitrogen) + 3 * 16.00 g/mol (oxygen) = 63.01 g/mol
Moles of HNO3 = 2.27 g HNO3 / 63.01 g/mol = 0.036 moles
Step 4: Determine the empirical formula.
To find the empirical formula, we need to determine the ratio of moles for each element.
The smallest number of moles among carbon, hydrogen, nitrogen, oxygen, and sulfur is 0.026 moles from the second sample for sulfur.
Dividing the number of moles of each element by 0.026 gives:
Carbon: 0.096 moles / 0.026 moles = 3.69
Hydrogen: 0.056 moles / 0.026 moles = 2.15
Nitrogen: 0.036 moles / 0.026 moles = 1.38
Considering these ratios, we can round the numbers to the nearest whole number:
Carbon: 4
Hydrogen: 5
Nitrogen: 1
Oxygen: ?
Step 5: Calculate the number of moles of oxygen.
The total number of moles of all elements in the compound can be calculated from the difference between the mass of the compound sample and the sum of masses of carbon, hydrogen, nitrogen, and sulfur products.
Mass of compound = mass of sample - mass of products
Mass of compound = 2.52 g - (4.23 g + 1.01 g + 2.11 g)
Mass of compound = 2.52 g - 7.35 g = -4.83 g
Since we have a negative mass of the compound, there is clearly an error in the calculations or the given information. Please double-check your numbers and correct any mistakes.
If you have any further questions, feel free to ask.