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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Riboflavin is one of the \(\mathrm{B}\) vitamins. It is also known as vitamin \(\mathrm{B}_{6}\) and is made up of carbon, hydrogen, nitrogen, and oxygen atoms. When \(10.00 \mathrm{~g}\) of vitamin \(\mathrm{B}_{6}\) is burned in oxygen, \(19.88 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(4.79 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are obtained. Another experiment shows that vitamin \(\mathrm{B}_{6}\) is made up of \(14.89 \% \mathrm{~N}\). What is the simplest formula for vitamin \(\mathrm{B}_{6}\) ?

Aluminum reacts with sulfur gas to form aluminum sulfide. Initially, \(1.18\) mol of aluminum and \(2.25\) mol of sulfur are combined. (a) Write a balanced equation for the reaction. (b) What is the limiting reactant? (c) What is the theoretical yield of aluminum sulfide in moles? (d) How many moles of excess reactant remain unreacted?

Epsom salts are hydrates of magnesium sulfate. The formula for Epsom salts is \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\). A \(7.834\) -g sample is heated until a constant mass is obtained indicating that all the water has been evaporated off. What is the mass of the anhydrous magnesium sulfate? What percentage of the hydrate is water?

Suppose that the atomic mass of \(\mathrm{C}-12\) is taken to be \(5.000 \mathrm{amu}\) and that a mole is defined as the number of atoms in \(5.000 \mathrm{~kg}\) of carbon-12. How many atoms would there be in one mole under these conditions? (Hint: There are \(6.022 \times 10^{23} \mathrm{C}\) atoms in \(12.00 \mathrm{~g}\) of \(\mathrm{C}-12 .\) )

Diborane, \(\mathrm{B}_{2} \mathrm{H}_{6}\), can be prepared by the following reaction: $$ 3 \mathrm{NaBH}_{4}(s)+4 \mathrm{BF}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{~B}_{2} \mathrm{H}_{6}(g)+3 \mathrm{NaBF}_{4}(s) $$ (a) How many moles of \(\mathrm{NaBH}_{4}\) react with \(1.299 \mathrm{~mol}\) of \(\mathrm{BF}_{3} ?\) (b) How many moles of \(\mathrm{B}_{2} \mathrm{H}_{6}\) can be obtained from \(0.893 \mathrm{~mol}\) of \(\mathrm{NaBH}_{4} ?\) (c) If \(1.987 \mathrm{~mol}\) of \(\mathrm{B}_{2} \mathrm{H}_{6}\) is obtained, how many moles of \(\mathrm{NaBF}_{4}\) are produced? (d) How many moles of \(\mathrm{BF}_{3}\) are required to produce \(4.992 \mathrm{~mol}\) of \(\mathrm{NaBF}_{4} ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free