Question

26) Which of the following is an empirical formula of a compound that is 52.15 percent C, 13.13 percent H, and 34.73 percent O? SHOW WORK 26) A) C4H1202 B) C2H6O C) CH602 D) C3H06 E) None of the answers is correct. 27) Calculate the formula mass of (NH4)3AsO4. A) 165.02 amu B) 156.96 amu C) 417.80 amu D) 108.96 amu E) 193.05 amu

Answer 1

Answer 26). C - 52.15% , H- 13.13% , O- 34.73%

step 1 -> devide these % with their mass.

C-> 52.15/12 = 4.34 | H-> 13.13/1 = 13.13 | O-> 34.73/16 = 2.17

Now divide these with lowest number 2.17

c-> 4.34/2.17 =2

H-> 13.13/2.17 = 6

O-> 2.17/2.17 = 1

number of C is 2, Number of H is 6 and number of O is 1

formula is - C₂H₆O

Answer 27) mass of individual atoms

Element	mass of one atom	number of atoms available in compound	total mass
N	14	3	42
H	1	12	12
As	75	1	75
O	16	6	64
total			193

Total mass is 193 amu