Question

determine empirical formula of compound with the following composition by mass:48.0%C,4.0%H,and 48.0%O.[C4H4O3]​

Answer 1

Glucose has empirical formula CH2O and relative molecular mass of 180. Find the molecular formula of glucose.(given: H = 1.0; C = 12.0; O = 16.0)Suggested Answer:Relative mass of Empirical Formula, CH2O = 12 + 2(1) + 16 = 30Relative molecular mass = n X relative mass of the empirical formulathus, n = Relative molecular mass / relative mass of the empirical formula= 180 / 30 = 6Therefore, Molecular Formula of Glucose = 6 X CH2O = C6H12O6Empirical and Molecular Formula QuestionsYou should be able to do these questions without a calculator.1. Determine the empirical formula of a compound with the followingcomposition by mass: 48.0 % C, 4.0 % H and 48.0 % O. [C4H4O3]2. Determine the empirical formula of a compound with the followingcomposition by mass: 36.0 % C, 4.0 % H, 28.0 % N and 32.0 % O.[C3H4N2O2]3. Determine the empirical formula of a compound with the followingcomposition by mass: 24.0 % C, 7.0 % H, 38.0 % F and 31.0 % P. [C2H7F2P]4. Determine the empirical formula of a compound with the followingcomposition by mass: 48.0 % C, 8.0 % H, 28.0 % N and 16.0 % O. If thiscompound has a molar mass of 200 g/mol , what is its molecularformula?[C4H8N2O and C8H16N4O2]5. A 100 mg sample of a compound containing C, H and O is burned and

Answer 2

The empirical formula for the given compound is CHO

Explanation:

We are given:

Percentage of C = 48 %

Percentage of H = 4 %

Percentage of O = 48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 48 g

Mass of H = 4 g

Mass of O = 48 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{48g}{12g/mole}=4moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{4g}{1g/mole}=4moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{48g}{16g/mole}=3moles$

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3 moles.

For Carbon = $\frac{4}{3}=1.33\approx 1$

For Hydrogen = $\frac{4}{3}=1.33\approx 1$

For Oxygen = $\frac{3}{3}=1$

• Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 1

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