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3 years ago

A compound has the formula X2O3 where x is an unknown element. If the compound is 24.26% oxygen, what element does x represent?

Chemistry

1 answer:

Vinil7 [7]3 years ago

5 0

<h3>Answer:</h3>

Element X is Arsenic (Ar)

<h3>Explanation:</h3>

We are given the compound X₂O₃

The compound is 24.26% Oxygen

We are required to identify element X

<h3>Step 1: Determine the mass of Oxygen in the compound </h3>

Atomic mass of Oxygen is 16.0 g

There are three oxygen atoms in the compound

Therefore;

Mass of oxygen in the compound = 3 × 16

= 48 g

<h3>Step 2: Total mass of the compound </h3>

Total percentage = 100%

But, Oxygen in the compound is 24.26 % and has a mass of 48 g

Therefore;

Mass of the compound = (48 g ÷ 24.26 ) × 100

= 197.857 g

<h3>Step 3 mass of X in the compound </h3>

Since we know the mass of oxygen in the compound and the mass of the compound we can determine the mass of X

Mass of X in the compound = Mass of the compound - Mass of Oxygen in the compound

= 197.857 g - 48 g

= 149.857 g

But there are two atoms of X in the compound.

Therefore, molar mass of X = 149.857 g ÷ 2

= 74.929 g

<h3>Step 4: Identity of element X</h3>

We then we need to identify the element with a molar mass of 74.929 g

Therefore, the element represented above is Arsenic which has an atomic mass of 74.92.

Thus, our compound is Ar₂O₃

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Answer:

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3 years ago

Find the empirical formula of the following compounds:

Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus $\frac{mol phosphorus}{ 30.97 g phosphorus}$ = 0.0293 mol

Moles bromine 6.99 g bromine $\frac{mol bromine}{79.90 g bromine}$ =0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

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8 0

1 year ago

1s22s22p63s23p64s23d104p65s24d105p5

Phantasy [73]

this element must be <span>iodine.</span>

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3 years ago

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