01:160:161 Chapter Notes - Chapter 4.11-4.12: Molar Mass, Combustion Analysis, Chemical Formula

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23 Oct 2018

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The empirical formula represents a ratio of atoms or a ratio of moles of atoms, not a ratio of masses. For the compound: we divide all the subscripts by the smallest one, once both elements have a whole number subscript, that is our empirical formula. We can determine the molecular formula of a compound from the empirical formula if we also know the molar mass of the compound molecular formula= empirical formula x n, where n= 1,2,3 . You divide the molar mass of the compound by the empirical formula molar mass in order to get n, and then multiply n by the empirical formula to get your molecular formula. In combustion analysis, the unknown compound undergoes combustion (or burning) in the presence of pure oxygen. When the sample is burned, all of the carbon in the sample is converted to co2, and all the hydrogen is converted to h2o. Combustion is a type of chemical reaction.

Related Questions

Determine the empirical and molecular formulas of each of the following substances:

(a) Styrene, a compound substance used to make Styrofoam^® cups and insulation, contains 92.3% C and 7.7% H by mass and has a molar mass of 104 g/mol.

(b) Caffeine, a stimulant found in coffee, contains 49.5% C, 5.15% H, 28.9% N, and 16.5% O by mass and has a molar mass of 195 g/mol.

(c) Monosodium glutamate (MSG), a flavor enhancer in certain foods, contains 35.51% C, 4.77% H, 37.85% O, 8.29% N, and 13.60% Na, and has a molar mass of 169 g/mol.

Empirical formula of this compound is CH.

To determine the molecular formula, we need to find out how many times is molar mass higher than the mass of empirical formula:

Molecular formula is C₈H₈.

b) Empirical formula of caffeine:

Empirical formula of caffeine is C₄H₅N₂O. The sum of atomic masses from empirical formula is 97 g/mol.

Molecular formula of caffeine is 2 × empirical formula = C₈H₁₀N₄O₂

c) Empirical formula of MSG:

Empirical formula of MSG is C₅H₈NO₄Na. The sum of atomic masses of empirical formula is 169 g/mol, the same as molar mass. Empirical and molecular formula are identical.

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1) How many atoms are there in 51.3 g of barium chloride (BaCl2)?
2) How many moles of carbon monoxide (CO) are in 29.3 g of the compound?

3) What is the formula for copper(II) phosphate? Capitalization and punctuation count?

4) How many atoms of phosphorus are in 2.90 mol of copper(II) phosphate?

5) How many moles of N are in 0.191 g of N2O?

6) A sample of C3H8 has 7.68 Ã 1024 H atoms. How many carbon atoms does the sample contain? What is the total mass of the sample?

7) Calculate the percent composition by mass of each element in Al(OH)3. Use at least three significant figures.

% Al= ?, % O=? , % H =?

8) What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass?

HCO (Insert subscripts as needed.)

9) A compound with the empirical formula CH2 has a molar mass of 98 g/mol. What is the molecular formula for this compound?

Answer Question 15
Answer Question 16

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01:160:161 Chapter Notes - Chapter 4.11-4.12: Molar Mass, Combustion Analysis, Chemical Formula

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