A compound has molar of 86 g/mol and has the percent composition (by mass) of a

Question

A compound has molar of 86 g/mol and has the percent composition (by mass) of a 55.8% C, 37.2% O, and 7.0% H. Determine the empirical formula: a. C H O_2 b. C H O c. C H O d. C H O Choose the pair of compounds with the same empirical formula. a. C_2 H_2 and C H b. NaHCO and Na_2CO_3 K_2 CrO_4 and K_2CR_2O_7 d. H_2O and H_2O_2 Calculate the molecular formula of a compound with the empirical formula CH_2O and a molar mass of 150 g/mol. a. C_2H O_2 b. C_3H_6O_3 c. C H_gO_4 d. C_5H_10O_5 In the commercial production of the element arsenic, arsenic(III) oxide is heated with carbon, which reduces the oxide to the metal according to the following equation: 2As_2O_3 + 3C rightarrow 3CO_2 + 4As If 8.87 of As_2O_3 is used in the reaction and 5.33 g of As is produced (actual mass), what is the percent yield? a 79.105% b. 18% c. 68, 53% d. 50.06%

Explanation / Answer

7) Given percent composition,

100 g of the compound contains 55.8 g C, 37.2 g O and 7.0 g H.

Atomic masses of the elements are: C = 12 g/mol; O = 16 g/mol and H = 1 g/mol.

Calculate the moles of the elements: C = (55.8 g)/(12 g/mol) = 4.65 mole; O = (37.2 g)/(16 g/mol) = 2.325; H = (7.00 g)/1.0 g/mol) = 7 mole.

Calculate the simple ratio of the moles of the elements: C:O:H = 4.65:2.235:7.00 = 2.080:1:3.131

2:1:3

The empirical formula of the compound is C2OH3.

Let the molecular formula be (C2OH3)n where n = integer.

The molar mass is (2*12 + 1*16 + 3*1)*n = 43n

Therefore, 43n = 86

===> n = 2

The molecular formula is C4O2H6 (ans).

8) The ratio of the number of atoms of the elements must be the same. Only for C2H2 and C6H6, the ratio C:H = 1:1; therefore, these two have the same empirical formula.

Ans: (a)

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