Need help on two questions (electron configuration and stoichiometry

[roxy6]

A compound is analyzed and it contains H,N,O and C. The original sample is 1.3957 grams. The sample is burned in excess oxygen to form 2.24 grams of CO2. 0.04 grams H is evaluated in the original sample.

a. What are the grams of oxygen in the original sample?
b. The original compound contains 28.84% N, how many grams of N are there?
c. How many grams of C are in the original compound?
d. What is the empirical formula for this compound?

How would you do this step by step and what are the solutions?

I was only able to get part B which was .0403 grams i believe?

I have a midterm tomorrow and need urgent help.

Thank you

 

[gravenewworld]

A compound is analyzed and it contains H,N,O and C. The original sample is 1.3957 grams. The sample is burned in excess oxygen to form 2.24 grams of CO2. 0.04 grams H is evaluated in the original sample.

a. What are the grams of oxygen in the original sample?
b. The original compound contains 28.84% N, how many grams of N are there?
c. How many grams of C are in the original compound?
d. What is the empirical formula for this compound?

How would you do this step by step and what are the solutions?

I was only able to get part B which was .0403 grams i believe?

I have a midterm tomorrow and need urgent help.

Thank you

b- correct
c- how many moles of C are in 2.24 grams of CO2? What is that in grams?
a- Subtract b, c, and 0.04 grams H from the original sample weight. That gives you the amount of oxygen in the original compound.
d-after you find everything in grams you can covert it to moles. Find the ratios between each government.

 

[Blueshift5]

for reaching out for assistance with your questions. I am happy to help guide you through the problem-solving process and provide solutions.

a. To determine the grams of oxygen in the original sample, we need to use stoichiometry. First, we need to determine the number of moles of CO2 formed in the reaction. We know that 2.24 grams of CO2 was formed, and the molar mass of CO2 is 44.01 g/mol. Using this information, we can calculate the number of moles of CO2 formed:
2.24 g CO2 x (1 mol CO2/44.01 g CO2) = 0.0509 mol CO2

Next, we need to use the balanced chemical equation for the combustion of the compound to determine the moles of oxygen consumed in the reaction. The balanced equation is:
C x H y N z O a + O2 → CO2 + H2O + N2

From the equation, we can see that for every 1 mole of CO2 formed, 1 mole of oxygen is consumed. Therefore, the number of moles of oxygen consumed is also 0.0509 mol.

Now, we can use the molar mass of oxygen (16.00 g/mol) to calculate the grams of oxygen in the original sample:
0.0509 mol O2 x (16.00 g O2/1 mol O2) = 0.8144 g O2

Therefore, there were 0.8144 grams of oxygen in the original sample.

b. As you correctly calculated, the grams of hydrogen in the original sample is 0.0403 g. To determine the grams of nitrogen in the original sample, we can use the percent composition given (28.84%). This means that 28.84% of the original sample is nitrogen, and the remaining 71.16% is made up of the other elements. We can set up the following equation to solve for the grams of nitrogen:
0.0403 g H x (100%/28.84%) = 0.1398 g N

Therefore, there were 0.1398 grams of nitrogen in the original sample.

c. To determine the grams of carbon in the original sample, we can use the same approach as in part b. The percent composition of carbon can be calculated by subtracting the percentages of the other elements