A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
) Maleic acid contains carbon, hydrogen and oxygen and has a molar mass of 116.08g/mol. When 10.0 g is combusted with excess oxygen, 15.2 g of carbon dioxide and 3.10 g of water are produced. Determine the molecular formula.
anonymous
 one year ago
) Maleic acid contains carbon, hydrogen and oxygen and has a molar mass of 116.08g/mol. When 10.0 g is combusted with excess oxygen, 15.2 g of carbon dioxide and 3.10 g of water are produced. Determine the molecular formula.

This Question is Open

matt101
 one year ago
Best ResponseYou've already chosen the best response.0Bear with me because this explanation is a bit long, but I want to explain the process as best as possible. Let me know if you need me to clarify anything further! We're assuming complete combustion here, and we know maleic acid must contain some combination of C's, H's, and O's. Write out a combustion reaction based on what you know from the question as follows: \[C_a H_b O_c + xO_2 \rightarrow aCO_2 + \frac{b}{2}H_2O\] The O's in CO2 and H2O might come from O2 or maleic acid, so we can't say much about the value of c. For the same reason, we can't figure out the value of x. However, all the C's in CO2 must come from maleic acid, so you have as many CO2's produced as there are C's in maleic acid. Similarly, all the H's in H2O must come from maleic acid, so you have HALF as many H2O's produced as there are H's in maleic acid. This is where the a and b/2 coefficients come from. You'll see we don't need to worry about the values of c or x. Keep your eye on the prize: maleic acid. We want to get the missing information  the amount of O it contains. We're given a mass of maleic acid, and we can figure out masses of C and H with what we're given to solve for the unknown  the mass of O. But, we're not given the masses of C and H; we're given the masses of CO2 and H2O. However, if we convert those masses to moles, and then consider the corresponding moles of just the atoms we're interested in, we'll be in business. Start with finding moles of CO2 and H2O: moles of CO2 = 15.2 g / 44 g/mol = 0.3454 mol moles of H2O = 3.10 g / 18 g/mol = 0.1722 mol Since we have 0.3454 mol of CO2, we have 0.3454 mol of C. Since we have 0.1722 mol of H2O, we have 0.3444 mol of H (it's doubled because every mole of H2O contains 2 moles of H; this accounts for the b/2 in the equation). We can use these numbers to find the mass of each element coming from maleic acid: mass of C = 0.3454 mol x 12 g/mol = 4.14 g mass of H = 0.3444 mol x 1 g/mol = 0.34 g The question told us we started with 10.0 g of maleic acid, and now we have masse of C and H, so we can solve for the remainder of the mass, which comes from O: 10.04.140.34 = 5.52 g. We can now figure out moles of O combusted by dividing by the molar mass of O: 5.52 g / 16 g/mol = 0.345 mol. We want to have moles of all three elements because now we can figure out the ratio of C, H, and O in maleic acid to come up with its empirical formula (divide by the lowest mole number to get a whole number ratio): 0.3454 C : 0.3444 H : 0.345 O = 1 C : 1 H : 1 O So the ratio is about 1:1:1, meaning the empirical formula is CHO. But we want the molecular formula  luckily we're given the molar mass of maleic acid, so we can figure this out. The mass of the empirical formula is 12 + 1 + 16 = 29 g/mol, while the mass of the maleic acid is 116.08 g/mol, which is about 4 times greater. This means there are actually 4 of each atom in maleic acid, making our molecular formula C4H4O4! Whew! Let me know if you have any questions!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.