A balloon contains 7.36 g of oxygen gas (02). How many oxygen molecules

Equilin is an estrogen isolated from the urine of pregnant mares. The percent composition of equilin is 80.56 percent C, 7.51 pe

tatuchka [14]

Answer:

The correct answer is C18H20O2.

Explanation:

Based on the given information, one hundred grams of equilin contains 80.56% Carbon, 7.51% Hydrogen, and 11.95% Oxygen. Thus, by mass it comprises 80.56 g Carbon, 7.51 g Hydrogen, and 11.92 g Oxygen.

The molecular mass of carbon, hydrogen, and oxygen are 12.01 g/mol, 1.008 g/mol, and 16.00 g/mol.

The moles of the elements present can be determined by using the formula,

Moles = Weight/Molecular mass

Moles of C = 80.56 g/12.01 g/mol = 6.708 mol

Moles of H = 7.51 g/1.008 g/mol = 7.45 mol

Moles of O = 11.92 g/16 g/mol = 0.745 mol

This gives the formula of C6.708H7.45O0.745. By dividing the subscripts from the smallest value, the possible whole numbers obtained will be,

6.708/0.745 = 9, 7.45/0.745 = 10, 0.745/0.745 = 1. Now the empirical formula obtained will be, C9H10O

Now dividing the molar mass of equilin, that is, 268 g/mol from the empirical formula mass, that is, 9*12.01 g/mol + 10*1.008 g/mol+ 16.00 g/mol = 134.2 g/mol, we get,

268/134.2 = 2

Now by multiplying the subscripts in the empirical formula with 2, the molecular formula obtained will be,

C18H20O2.

6 0

3 years ago

Can u guys please answer this<br><br> What plasma looks like on the molecular level?

meriva

They look like gases plasmas have no fixed shapes or volume and are less dense tan solids or liquids

5 0

3 years ago

Read 2 more answers

A sample of an unknown gas effuses in 14.4 min. An equal volume of H2 in the same apparatus under the same conditions effuses in

Elena-2011 [213]

Answer:- molar mass of the unknown gas is 71.5 gram per mol.

Solution:- From Graham's law of effusion rates, the rate of effusion of a gas is inversely proportional to the square root of it's molar mass.

When we compare the effusion rates of two gases then the formula for Graham's law is:

$\frac{rate_1}{rate_2}=\sqrt{\frac{M_2}{M_1}}$

In this formula, V stands for volume and M stands for molar mass

Rate is volume effused per unit time. Since, the volumes are same, the formula could be written as:

$\frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}$

let's say in formula, subscript 1 is for hydrogen gas and 2 is for the unknown gas.

Molar mass of hydrogen is 2.02 grams per mol and the time taken to effuse it is 2.42 min. The time taken to effuse the unknown gas is 14.4 min and we are asked to calculate it's molar mass. let's plug in the values in the formula:

$\frac{14.4}{2.42}=\sqrt{\frac{M_2}{2.02}}$