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What is the empirical formula for a compound that is 43.6% phosphorus
and 56.4% oxygen?
Select one:
O a. P307
O b. PO3
O C. P203
O d. P205
1727 of what is the empirical

Respuesta :

Answer:

d.   P2O5.

Explanation:

We find the ratio of the atoms by dividing the percentages by the relative atomic masses:

P :  43.6 / 30.974 = 1.4076

O:  56.4 / 15.999 =  3.5252

1.4076 : 3.5252

= 1 : 2.5

= 2:5.

So the answer is P2O5.

samueladesida43

The empirical formula for a compound that is 43.6% phosphorus and 56.4% oxygen is P2O5.

HOW TO CALCULATE EMPIRICAL FORMULA:

The empirical formula of a compound can be calculated using the following process:

  • P = 43.6% = 43.6g
  • O = 56.4% = 56.4g

First, we divide by the molar mass of each element as follows:

  • P = 43.6g ÷ 31g/mol = 1.41mol
  • O = 56.4g ÷ 16g/mol = 3.53mol

Next, we divide each mole value by the smallest

  • P = 1.41mol ÷ 1.41 = 1
  • O = 3.53mol ÷ 1.41 = 2.5

We multiply each ratio by 2 to get an approximate value of 2:5 for P and O respectively.

Therefore, the empirical formula for a compound that is 43.6% phosphorus and 56.4% oxygen is P2O5.

Learn more about empirical formula at: https://brainly.com/question/11588623

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