A compound contains only carbon, hydrogen, and oxygen. Combustion of 65.76 g of the compound yields 96.38 g of CO2 and 39.46 g of H2O. The molar mass of the compound is 90.078 g/mol.
1. Calculate the grams of carbon (C) in 65.76 g of the compound:
2. Calculate the grams of hydrogen (H) in 65.76 g of the compound.
3. Calculate the grams of oxygen (O) in 65.76 g of the compound.

1. 26.30 g C.

2. 4.42 g H.

3. 35.04 g O.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required as follows:

1. Here, the only source of carbon is in CO2, and thus, we calculate the grams of carbon from the produced grams of this substance:

2. Here, the only source of hydrogen is in H2O, and thus, we calculate the grams of hydrogen from the produced grams of this substance:

3. Here, we subtract the mass of H and C from the mass of the sample, to obtain the mass of oxygen:

Regards!

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years