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Chapter 3: Problem 47

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{CH}_{2}\), molar mass \(=84 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{NH}_{2} \mathrm{Cl}\), molar mass \(=51.5 \mathrm{~g} / \mathrm{mol}\)

Short Answer

Expert verified
The molecular formula for the compounds are: (a) \(\mathrm{C}_{6} \mathrm{H}_{12}\) (b) \(\mathrm{NH}_{2}\mathrm{Cl}\)

Step by step solution

01

(a) Step 1: Calculate the empirical formula weight for CH2

Calculate the empirical formula weight by adding the atomic weights of the elements in the empirical formula. For \(\mathrm{CH}_{2}\): Empirical Formula Weight = Atomic Weight of Carbon + (2 脳 Atomic Weight of Hydrogen) = 12.01 + (2 脳 1.01) = 12.01 + 2.02 = 14.03 g/mol
02

(a) Step 2: Find the ratio between molar mass and empirical formula weight

Divide the molar mass given by the empirical formula weight. Ratio = (Molar Mass) / (Empirical Formula Weight) = 84 / 14.03 = 6
03

(a) Step 3: Determine the molecular formula

Multiply each element in the empirical formula by the ratio found in Step 2. Molecular Formula = \(\mathrm{C}(6) \mathrm{H}_{2}(6) = \mathrm{C}_{6} \mathrm{H}_{12}\) So, the molecular formula for a compound with an empirical formula of \(\mathrm{CH}_{2}\) and molar mass of 84 g/mol is \(\mathrm{C}_{6} \mathrm{H}_{12}\).
04

(b) Step 1: Calculate the empirical formula weight for NH2Cl

Calculate the empirical formula weight by adding the atomic weights of the elements in the empirical formula. For \(\mathrm{NH}_{2} \mathrm{Cl}\): Empirical Formula Weight = Atomic Weight of Nitrogen + (2 脳 Atomic Weight of Hydrogen) + Atomic Weight of Chlorine = 14.01 + (2 脳 1.01) + 35.45 = 14.01 + 2.02 + 35.45 = 51.48 g/mol
05

(b) Step 2: Find the ratio between molar mass and empirical formula weight

Divide the molar mass given by the empirical formula weight. Ratio = (Molar Mass) / (Empirical Formula Weight) = 51.5 / 51.48 鈮� 1
06

(b) Step 3: Determine the molecular formula

Multiply each element in the empirical formula by the ratio found in Step 2. Since the ratio is approximately 1, the molecular formula is the same as the empirical formula. Molecular Formula = \(\mathrm{NH}_{2}\mathrm{Cl}\) So, the molecular formula for a compound with an empirical formula of \(\mathrm{NH}_{2} \mathrm{Cl}\) and molar mass of 51.5 g/mol is \(\mathrm{NH}_{2}\mathrm{Cl}\).

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Most popular questions from this chapter

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of \(\mathrm{CO}, 0.733 \mathrm{~g}\) of \(\mathrm{CO}_{2}\), and \(0.450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

The fermentation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) produces ethyl alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(\mathrm{CO}_{2}\) : $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ (a) How many moles of \(\mathrm{CO}_{2}\) are produced when \(0.400\) mol of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) reacts in this fashion? (b) How many grams of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) ? (c) How many grams of \(\mathrm{CO}_{2}\) form when \(7.50 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) are produced?

Calcium hydride reacts with water to form calcium hydroxide and hydrogen gas. (a) Write a balanced chemical equation for the reaction. (b) How many grams of calcium hydride are needed to form \(8.500 \mathrm{~g}\) of hydrogen?

(a) What is the difference between adding a subscript 2 to the end of the formula for \(\mathrm{CO}\) to give \(\mathrm{CO}_{2}\) and adding a coefficient in front of the formula to give 2 CO? (b) Is the following chemical equation, as written, consistent with the law of conservation of mass? \(3 \mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{~s})+2 \mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l)\) Why or why not?

Several brands of antacids use \(\mathrm{Al}(\mathrm{OH})_{3}\) to react with stomach acid, which contains primarily HCl: $$ \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ (a) Balance this equation. (b) Calculate the number of grams of \(\mathrm{HCl}\) that can react with \(0.500 \mathrm{~g}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\). (c) Calculate the number of grams of \(\mathrm{AlCl}_{3}\) and the number of grams of \(\mathrm{H}_{2} \mathrm{O}\) formed when \(0.500 \mathrm{~g}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.
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