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How many moles of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O},\) are in a \(10.0 \mathrm{~g}\) sample? How many grams are in a \(0.10 \mathrm{~mol}\) sample of ethanol?

Short Answer

Expert verified
10.0 g of ethanol is 0.217 moles, and 0.10 moles is 4.608 g of ethanol.

Step by step solution

01

Determine Molar Mass of Ethanol

Ethanol, with the chemical formula \( \mathrm{C}_{2}\mathrm{H}_6\mathrm{O} \), consists of 2 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom. Use the atomic masses from the periodic table: Carbon (C) = 12.01 g/mol, Hydrogen (H) = 1.01 g/mol, and Oxygen (O) = 16.00 g/mol. Calculate the molar mass as follows:\[\text{Molar Mass of } \mathrm{C}_2\mathrm{H}_6\mathrm{O} = 2(12.01) + 6(1.01) + 16.00 = 46.08 \,\mathrm{g/mol}.\]
02

Convert Grams to Moles

To convert the given mass of ethanol (10.0 g) to moles, use the molar mass calculated in Step 1. The formula is:\[\text{Moles of Ethanol} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{10.0\,\text{g}}{46.08\,\text{g/mol}}.\]Calculating gives:\[0.217\,\text{moles of ethanol}.\]
03

Convert Moles to Grams

To find the mass of a 0.10 mol sample of ethanol, use the molar mass determined in Step 1:\[\text{Mass of Ethanol} = \text{Moles} \times \text{Molar Mass} = 0.10\,\text{mol} \times 46.08\,\text{g/mol}.\]Calculating gives:\[4.608\,\text{g of ethanol}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
Understanding molar mass is key to any mole calculation. A substance's molar mass connects its mass with the amount of substance it contains in moles. Moles are a unit like dozens or pairs, but used for atoms and molecules.
Each substance has a characteristic molar mass, represented in grams per mole (g/mol). To find the molar mass, sum up the atomic masses of all the atoms in the molecule using a periodic table.
For ethanol (\(\mathrm{C}_{2}\mathrm{H}_6\mathrm{O}\)), you add:
  • 2 Carbon atoms (C), each with an atomic mass of 12.01 g/mol, giving a total of \(2 \times 12.01 = 24.02\,\mathrm{g/mol}\)
  • 6 Hydrogen atoms (H), each with an atomic mass of 1.01 g/mol, for a total of \(6 \times 1.01 = 6.06\,\mathrm{g/mol}\)
  • 1 Oxygen atom (O), with an atomic mass of 16.00 g/mol
Adding these gives ethanol a molar mass of 46.08 g/mol. This value allows conversions between grams and moles for this substance. By knowing the molar mass, you can solve several chemical problems.
Convert Grams to Moles
Converting grams to moles is a process that allows us to understand how many moles are present in a given mass of a substance. To do this conversion, use the formula: \[\text{Moles} = \frac{\text{Grams}}{\text{Molar Mass}}.\]For example, if you have a 10.0 g sample of ethanol with a molar mass of 46.08 g/mol, you calculate the moles as follows:
  • Plug into the formula: \(\text{Moles} = \frac{10.0 \, \text{g}}{46.08 \, \text{g/mol}}\)
  • Do the division: \(10.0\/46.08\approx{0.217} \text{ moles}\)
So, the 10.0 g sample of ethanol is roughly 0.217 moles. This step is a fundamental part of chemistry calculations where comparing amounts of substances is crucial.
Convert Moles to Grams
Going from moles to grams allows us to find the mass of a substance for a given number of moles. This reverse calculation is straightforward when you know the molar mass. Use the formula:
\[\text{Grams} = \text{Moles} \times \text{Molar Mass}.\]Take a 0.10 mol sample of ethanol as an example:
  • Use the formula: \(\text{Grams} = 0.10 \, \text{mol} \times 46.08 \, \text{g/mol}\)
  • Multiply: \(0.10 \times 46.08 = 4.608 \, \text{g}\)
This tells you that 0.10 moles of ethanol weighs 4.608 grams. Whether you're working in a lab or tackling homework essays, mastering these conversions is essential for accurately measuring chemicals.

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Most popular questions from this chapter

The net ionic equation for the Breathalyzer test used to indicate alcohol concentration in the body is as follows: \(16 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+3 \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}(a q) \longrightarrow\) $$3 \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}(a q)+4 \mathrm{Cr}^{3+}(a q)+11 \mathrm{H}_{2} \mathrm{O}(l)$$ (a) How many grams of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) must be used to consume \(1.50 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O} ?\) (b) How many grams of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) can be produced from \(80.0 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O} ?\)

Dichloromethane, \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\), the solvent used to decaffeinate coffee beans, is prepared by reaction of \(\mathrm{CH}_{4}\) with \(\mathrm{Cl}_{2}\). (a) Write the balanced equation. (HCl is also formed.) (b) How many grams of \(\mathrm{Cl}_{2}\) are needed to react with \(50.0 \mathrm{~g}\) of \(\mathrm{CH}_{4} ?\) (c) How many grams of dichloromethane are formed from \(50.0 \mathrm{~g}\) of \(\mathrm{CH}_{4}\) if the percent yield for the reaction is \(76 \% ?\)

At elevated temperatures in an automobile engine, \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) can react to yield NO, an important cause of air pollution. (a) Write a balanced equation for the reaction. (b) How many moles of \(\mathrm{N}_{2}\) are needed to react with \(7.50 \mathrm{~mol}\) of \(\mathrm{O}_{2} ?\) (c) How many moles of \(\mathrm{NO}\) can be formed when \(3.81 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) reacts? (d) How many moles of \(\mathrm{O}_{2}\) must react to produce \(0.250 \mathrm{~mol}\) of \(\mathrm{NO} ?\)

The active ingredient in milk of magnesia (an antacid) is magnesium hydroxide, \(\mathrm{Mg}(\mathrm{OH})_{2}\). A typical dose (one tablespoon) contains \(1.2 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} .\) Calculate (a) the molar mass of magnesium hydroxide and (b) the amount of magnesium hydroxide (in moles) in one tablespoon.

In Problem 6.40 , hydrazine reacted with oxygen according to the following (unbalanced) equation: $$\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ (a) If \(75.0 \mathrm{~kg}\) of hydrazine are reacted with \(75.0 \mathrm{~kg}\) of oxygen, which is the limiting reagent? (b) How many kilograms of \(\mathrm{NO}_{2}\) are produced from the reaction of \(75.0 \mathrm{~kg}\) of the limiting reagent? (c) If \(59.3 \mathrm{~kg}\) of \(\mathrm{NO}_{2}\) are obtained from the reaction in part (a), what is the percent yield?

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