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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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Key Concepts

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

Chemical Calculations
Chemical calculations are a fundamental part of chemistry, allowing us to predict amounts of reactants or products in a given reaction. In this problem, chemical calculations help us determine the mass of sodium benzoate in a specific amount of salami. We start by finding the percentage of sodium benzoate in salami (0.090%) and use it to calculate the exact mass present when consuming 6 ounces of salami, ultimately converting units to work in grams since scientific calculations often use SI units.
  • First, the salami's weight in ounces is converted to grams. This ensures that subsequent calculations regarding concentration and mass are consistent.
  • By determining how much sodium benzoate is present in those grams of salami, we see practical application of percentages, highlighting how a seemingly small concentration can significantly impact calculations.
These steps are the backbone of any chemical analysis and showcase fundamental skills needed for more complex calculations.
Molar Mass
Molar mass is the mass of one mole of a compound. It helps chemists convert between mass and moles, crucial for determining the number of molecules. In this problem, we calculate the molar mass of sodium benzoate (\(\mathrm{NaC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\)), a common preservative.
Calculating the molar mass involves summing the atomic masses of all atoms in a molecule. For sodium benzoate:
  • Sodium (Na): 22.99 g/mol
  • Carbon (C): 12.01 g/mol, multiplied by 7
  • Hydrogen (H): 1.01 g/mol, multiplied by 5
  • Oxygen (O): 16.00 g/mol, multiplied by 2
Adding these gives a molar mass around 144.11 g/mol. Understanding molar mass is essential for converting grams to moles, as gram measurements need to be expressed in the more manageable concept of moles for reactions.
Avogadro's Number
Avogadro's number is a key concept in chemistry, defined as the number of units (atoms, molecules) in one mole of a given substance. Here, it helps us convert moles of sodium to the number of atoms.
Avogadro's number (\(N_A\)) is approximately \(6.022 \times 10^{23}\) atoms/mol. With this constant:
  • We can translate a mole-based measurement (such as moles of sodium atoms) into an actual count of atoms, useful for understanding reactions on both a macroscopic and molecular level.
  • By multiplying the moles of sodium we calculated by Avogadro's number, we determine the immense quantity of atoms you consume — illustrating the concept that even small amounts of a compound involve enormous numbers of atoms.
This conversion helps illustrate the fundamental, vast scale at which chemical reactions and compositions occur.
Conversion of Units
Converting units is a daily practice in chemistry and is fundamental to solving real-world problems. Here, understanding conversions was crucial for transitioning from ounces to grams, ensuring the correct use of SI units in calculations.
  • In this problem, converting 6 ounces of salami into grams (\(6 \text{ oz} \times 28.35 \text{ g/oz} = 170.1 \text{ g}\)) was essential before any other calculation could proceed.
  • Appropriate unit conversion is also vital when determining the percentage mass (as percentages need consistent base units), ensuring that all subsequent calculations derive from a common measurement.
Proficiency in conversion helps prevent errors and ensures accuracy, fundamental to both laboratory work and in understanding complex chemical processes.

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Most popular questions from this chapter

Most wine is prepared by the fermentation of the glucose in grape juice by yeast: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ How many grams of glucose should there be in grape juice to produce \(725 \mathrm{~mL}\) of wine that is \(11.0 \%\) ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\left(d=0.789 \mathrm{~g} / \mathrm{cm}^{3}\right)\), by volume?

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.

Write balanced equations for the reaction of sulfur with the following metals to form solids that you can take to be ionic when the anion is \(\mathrm{S}^{2-}\). (a) potassium (b) magnesium (c) aluminum (d) calcium (e) iron (forming \(\mathrm{Fe}^{2+}\) ions)

Oxyacetylene torches used for welding reach temperatures near \(2000^{\circ} \mathrm{C}\). The reaction involved in the combustion of acetylene is $$ 2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) Starting with \(175 \mathrm{~g}\) of both acetylene and oxygen, what is the theoretical yield, in grams, of carbon dioxide? (b) If \(68.5 \mathrm{~L}(d=1.85 \mathrm{~g} / \mathrm{L})\) of carbon dioxide is produced, what is the percent yield at the same conditions of temperature and pressure? (c) How much of the reactant in excess is unused? (Assume 100\% yield.)

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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