Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen.

in an experiment, the molar mass of the compound was determined to be 228.276 g/mol. what is the molecular formula of the compound?

for both questions, show your work or explain how you determined the formulas by giving specific values used in calculations.

Hello!

The answer is:

Empirical formula= C3H8O2

Molecular formula= C9H24O6

Why?

First, we need to calculate how many moles of each element are present in the chemical reaction by dividing the given information (grams) into the molecular mass of each element:

Carbon: 12,011 g/mol

Hydrogen: 1,008 g/mol

Oxygen: 15,999 g/mol

Calculating the moles of each element:

Moles of Carbon =

Moles of Hydrogen =

Moles of Oxygen =

Then, we need to determinate the mole ratio. We can do it by dividing each result into the smallest value of moles (2.62 moles (O))

To get entire numbers, we will use the 2 factor,

So,the empirical  formula will be:

C3H8O2

To know the molecular formula,

We need to divide the given molar mass of the compound by the empirical formula molecular mass to find the ratio:

So, the molecular formula will be:

Have a nice day!

Hello!

Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen.  

In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?

We have the following data:

Carbon (C) ≈ 12 a.m.u (g/mol)

Hydrogen (H) ≈ 1 a.m.u (g/mol)

Oxygen (O) ≈ 16 a.m.u (g/mol)

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

C: 47.37 g

H: 10.59 g

O: 42.04 g

The values ​​(in g) will be converted into quantity of substance (number of mols), dividing by molecular weight (g / mol) each of the values, we will see:We realize that the values ​​found above are not integers, so we divide these values ​​by the smallest of them, so that the proportion does not change, let us see:

convert number of atomic radio into whole number

2 * (1.5 : 4 : 1)  

= 3 : 8 : 2  ← whole number of atomic radio

C = 3

H =  8

O = 2

Thus, the empirical formula found for the compound will be:We are going to find the Molar Mass (MM) of the Empirical Formula (EF), let's see:

if: C3H8O2

C = 3*(12 a.m.u) = 36 a.m.u  

H = 8*(1 a.m.u) = 8 a.m.u  

O = 2*(16 a.m.u) = 32 a.m.u  

Knowing that the Molar Mass of the Molecular Formula is 228.276 (in g/mol) and that the Molar Mass of the Empirical Formula is 76 (in g/mol), then we will find the number of terms (n) for the molecular formula of the compound, let us see:The Molecular Formula is the Empirical Formula times the number of terms (n), then, we have:

Answers:

The empirical formula: C3H8O2

The molecular formula: C9H24O6

1.39 × 10²³ particles CuCr₂O₇

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Chemistry

Atomic Structure

Reading a Periodic TableMolesAvogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional AnalysisExplanation:

Step 1: Define

[Given] 64.5 g CuCr₂O₇

[Solve] particles CuCr₂O₇

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Cu - 63.55 g/mol

[PT] Molar Mass of Cr - 52.00 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of CuCr₂O₇ - 63.55 + 2(52.00) + 7(16.00) = 279.55 g/mol

Step 3: Convert

[DA] Set up:                                                                                                     [DA] Divide/Multiply [Cancel out units]:                                                        

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.38944 × 10²³ particles CuCr₂O₇ ≈ 1.39 × 10²³ particles CuCr₂O₇