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Chapter 3: Problem 85

Some brands of salami contain \(0.090 \%\) sodium benzoate \(\left(\mathrm{NaC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\right)\) by mass as a preservative. If you eat \(6.00 \mathrm{oz}\) of this salami, how many atoms of sodium will you consume, assuming salami contains no other source of that element?

Short Answer

Expert verified
Answer: Approximately 6.39 × 10^{20} sodium atoms.

Step by step solution

01

Find the mass of sodium benzoate consumed

First, we need to find out how much sodium benzoate is present in the given amount of salami. To do this, we will multiply the percentage of sodium benzoate in the salami by the consumed mass. Given: Percentage of sodium benzoate in salami = \(0.090 \%\) Amount of salami consumed = \(6.00 \ oz\) First, we need to convert ounces to grams, as SI units are used in calculations. 1 ounce = 28.35 grams 6 oz × 28.35 g/oz ≈ 170.1 g Now we can find the mass of sodium benzoate consumed: Mass of sodium benzoate = (0.090/100) × 170.1 g ≈ 0.15309 g
02

Calculate the moles of sodium benzoate consumed

In order to find the moles of sodium benzoate, we need to divide the mass of consumed sodium benzoate by its molar mass. Molar mass of sodium benzoate (\(\mathrm{NaC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\)) = 22.99 g/mol (Na) + 12.01 g/mol × 7 (C) + 1.01 g/mol × 5 (H) + 16.00 g/mol × 2 (O) ≈ 144.11 g/mol Moles of sodium benzoate = (Mass of sodium benzoate) / (Molar mass of sodium benzoate) Moles of sodium benzoate ≈ 0.15309 g / 144.11 g/mol ≈ 0.001062 mol
03

Determine the moles of sodium atoms in the consumed sodium benzoate

In one mole of sodium benzoate, there is one mole of sodium atoms. Therefore, the moles of sodium atoms in the consumed sodium benzoate are equal to the moles of sodium benzoate consumed. Moles of sodium atoms = Moles of sodium benzoate ≈ 0.001062 mol
04

Convert the moles of sodium atoms to the number of sodium atoms

Finally, we can use Avogadro's number to convert the moles of sodium atoms to the number of sodium atoms. Avogadro's Number, \(N_{A} = 6.022 \times 10^{23}\) atoms/mol Number of sodium atoms = Moles of sodium atoms × Avogadro's Number Number of sodium atoms ≈ 0.001062 mol × (6.022 × 10^{23} atoms/mol) ≈ 6.39 × 10^{20} atoms So, if you eat 6.00 ounces of this salami, you will consume approximately 6.39 × 10^{20} sodium atoms.

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Most popular questions from this chapter

Determine whether the statements given below are true or false. (a) The mass of an atom can have the unit mole. (b) In \(\mathrm{N}_{2} \mathrm{O}_{4}\), the mass of the oxygen is twice that of the nitrogen. (c) One mole of chlorine atoms has a mass of \(35.45 \mathrm{~g}\). (d) Boron has an average atomic mass of \(10.81\) amu. It has two isotopes, \(\mathrm{B}-10(10.01\) amu \()\) and \(\mathrm{B}-11(11.01 \mathrm{amu}) .\) There is more naturally occurring B-10 than B-11. (e) The compound \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2} \mathrm{~N}\) has for its simplest formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{ON}_{1 / 2}\). (f) A 558.5-g sample of iron contains ten times as many atoms as \(0.5200 \mathrm{~g}\) of chromium. (g) If \(1.00\) mol of ammonia is mixed with \(1.00\) mol of oxygen the following reaction occurs, $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ All the oxygen is consumed. (h) When balancing an equation, the total number of moles of reactant molecules must equal the total number of moles of product molecules.

Phosphorus reacts with oxygen to produce different kinds of oxides. One of these oxides is formed when \(1.347 \mathrm{~g}\) of phosphorus reacts with \(1.744 \mathrm{~g}\) of oxygen. What is the simplest formula of this oxide? Name the oxide.

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Calculate the molar masses (in grams per mole) of (a) cane sugar, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) (b) laughing gas, \(\mathrm{N}_{2} \mathrm{O}\). (c) vitamin A, \(\mathrm{C}_{20} \mathrm{H}_{30} \mathrm{O}\).

Consider the hypothetical reaction $$ 8 \mathrm{~A}_{2} \mathrm{~B}_{3}(s)+3 \mathrm{X}_{4}(g) \longrightarrow 4 \mathrm{~A}_{4} \mathrm{X}_{3}(s)+12 \mathrm{~B}_{2}(g) $$ When \(10.0 \mathrm{~g}\) of \(\mathrm{A}_{2} \mathrm{~B}_{3}(\mathrm{MM}=255 \mathrm{~g} / \mathrm{mol})\) react with an excess of \(\mathrm{X}_{4}, 4.00 \mathrm{~g}\) of \(\mathrm{A}_{4} \mathrm{X}_{3}\) are produced. (a) How many moles of \(A_{4} X_{3}\) are produced? (b) What is the molar mass of \(\mathrm{A}_{4} \mathrm{X}_{3}\) ?
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