44 g CO2 contain 12 g C
11.085 g CO2 contain 12×11.085/44 g C
= 3.023 g C
18 g H2O contain 2 g H
2.2785 g H2O contain 2×2.2785/18 g H
= 0.2532 g H
Mass % of c = mass of C×100/total mass
= 3.023×100/3.2798
= 92.17%
Mass % of H = 100- 92.17 = 7.83%
Moles of C = mass/molar mass = 92.17/12 = 7.68 moles
Moles of H = 7.83/1 = 7.83 moles
Molar ratio = C : H = 7.68 : 7.83 = 1 : 1
Answer is C 9H9
Question 13 (4 points) In a quantitative analysis a 3.2798 g sample of a hydrocarbon was...
a combustion analysis of a 0.2104 g hydrocarbon sample (containing only hydrogen and carbon) yielded 0.6373 g carbon dioxide and 0.3259 g water. A separate experiment determined that the molecular weight of this compound is 58.7 g mol-1. from this data, determine both emperical and the molecular formula of this substance.
Fill in the Blanks A 27.0 g sample of an unknown hydrocarbon was burned in excess oxygen to form 88.0 g of carbon dioxide and 27 g of water. How many carbon and hydrogen atoms make up the empirical formula of this hydrocarbon? C atoms: Hatoms:
A hydrocarbon undergoes combustion analysis, and 4.40 g carbon dioxide and 2.70 g water vapor is obtained. What is the empirical formula of the hydrocarbon? (5 points) What is the difference between an empirical and molecular formula, including an example? (2 points)
2) A 1.000 g sample of a hydrocarbon (which is a compound that contains only carbon and hydrogen) was burned in oxygen, and the carbon dioxide and water produced were carefully collected and weighed. The mass of the carbon dioxide was 3.385 g, and the mass of the water was 0.692 g. What was the empirical formula of the hydrocarbon? A. C5H12 B. CsH6 C. CįH6 D. C H2 E. CįH star 1.00
12. What is the mass of 0.350 moles of molecular hydrogen gas (6 points) 13. How many atoms are in 0.027 moles KCl (6 points)? 14. In the body, the amino acid glycine (C2H5NO:) decomposes to carbon dioxide, water, and urea (CHAN;0). 2 GHNO, (aq) + 3 0,(g) 300,(g) + 3 H20 (1) + CH, N20 (aq) a. How many grams of urea are produced from 15.0 g glycine and an excess amount of oxygen (6 points)? b. How many...
15. The molecular formula of a gaseous hydrocarbon can be determined by combusting it completely in excess oxygen and then passing it through potassium hydroxide solution to absorb the carbon dioxide produced. In an experiment 200 cm' of a hydrocarbon was reacted with 1500 cm of oxygen. After the hydrocarbon had combusted completely 1000 cm of gas remained. This volume was reduced to 200 cm3 after the gas had been passed through a solution of potassium hydroxide. All volumes were...
6.0 grams hydrocarbon sample were made to react with oxygen gas to produce 17.6 grams of carbon dioxide according to the equation: CxHy(g) + O2(g) ---------- CO2(g) + H20(l) where x = subscript for C, y = subscript for H a. How many grams of C are there in 17.6 grams of CO2? b. How many grams of C are there in the hydrocarbon sample? c. How many moles of C are there in the hydrocarbon sample? d. How many...
35a When 4.752 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 16.06 grams of CO2 and 3.289 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 26.04 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon. 35 part b A 4.079 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 9.273 grams of CO2 and...
Furnace HO absorber CO, absorber Sample When 1.690 grams of a hydrocarbon, CyHy, were burned in a combustion analysis apparatus, 5.303 grams of Co, and 2.171 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 70.13 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon. Enter the elements in the order presented in the question. empirical formula molecular formula <Previous
1. A sample of a hydrocarbon weighing 4.99 grams was burned completely in excess oxygen, and 15.00 grams of carbon dioxide were obtained. What is the empirical formula of the compound? CHs CH4 О сн C3Hs