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Chapter 3: Problem 41

A sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) contains \(1.250 \times 10^{21}\) carbon atoms. (a) How many atoms of hydrogen does it contain? (b) How many molecules of glucose does it contain? (c) How many moles of glucose does it contain? (d) What is the mass of this sample in grams?

Short Answer

Expert verified
(a) The sample contains \(2.50 \times 10^{21}\) hydrogen atoms. (b) The sample contains \(2.083 \times 10^{20}\) glucose molecules. (c) The sample contains \(3.46 \times 10^{-4}\) moles of glucose. (d) The mass of the sample is approximately \(0.0623\) grams.

Step by step solution

01

Finding the number of hydrogen atoms

We know that a glucose molecule contains 6 carbon atoms and 12 hydrogen atoms. Since we are given the number of carbon atoms in the sample, we can use the ratio of hydrogen to carbon atoms in glucose to find the number of hydrogen atoms in the sample. Number of hydrogen atoms = (Number of carbon atoms in the sample * Number of hydrogen atoms in glucose) / Number of carbon atoms in glucose Number of hydrogen atoms = \((1.250 \times 10^{21} \; \text{carbon atoms}) * (12 \, \text{hydrogen atoms}) / (6 \, \text{carbon atoms})\)
02

Calculate the number of hydrogen atoms

Now let's simplify the expression from Step 1: Number of hydrogen atoms = \((1.250 \times 10^{21}) \times \frac{12}{6}\) Number of hydrogen atoms = \(1.250 \times 10^{21} \times 2\) Number of hydrogen atoms = \(2.50 \times 10^{21}\) So, the sample contains \(2.50 \times 10^{21}\) hydrogen atoms.
03

Finding the number of glucose molecules

Since we know each glucose molecule contains 6 carbon atoms, we can find the number of glucose molecules in the sample by dividing the total number of carbon atoms by the 6 carbon atoms in a glucose molecule: Number of glucose molecules = Total number of carbon atoms / Number of carbon atoms in glucose Number of glucose molecules = \((1.250 \times 10^{21} \, \text{carbon atoms}) / (6 \, \text{carbon atoms})\)
04

Calculate the number of glucose molecules

Now let's simplify the expression from Step 3: Number of glucose molecules = \(1.250 \times 10^{21} \, \text{carbon atoms} / 6\) Number of glucose molecules = \(2.083 \times 10^{20}\) So, the sample contains \(2.083 \times 10^{20}\) glucose molecules.
05

Finding the number of moles of glucose

To find the number of moles of glucose in the sample, we can divide the number of glucose molecules by Avogadro's number, which is approximately \(6.022 \times 10^{23} \, \text{molecules/mol}\): Number of moles of glucose = Number of glucose molecules / Avogadro's number Number of moles of glucose = \((2.083 \times 10^{20} \, \text{molecules}) / (6.022 \times 10^{23} \, \text{molecules/mol})\)
06

Calculate the number of moles of glucose

Now let's simplify the expression from Step 5: Number of moles of glucose = \(2.083 \times 10^{20} / 6.022 \times 10^{23}\) Number of moles of glucose = \(3.46 \times 10^{-4}\) mol So, the sample contains \(3.46 \times 10^{-4}\) moles of glucose.
07

Finding the mass of the sample in grams

To find the mass of the sample in grams, we can multiply the number of moles of glucose by the molar mass of glucose. The molar mass of glucose is approximately \(180.16 \, \text{g/mol}\): Mass of the sample = Number of moles of glucose * Molar mass of glucose Mass of the sample = \((3.46 \times 10^{-4} \, \text{mol}) * (180.16 \, \text{g/mol})\)
08

Calculate the mass of the sample in grams

Now let's simplify the expression from Step 7: Mass of the sample = \(3.46 \times 10^{-4} * 180.16\) Mass of the sample = \(0.0623 \, \text{g}\) So, the mass of the sample is approximately \(0.0623\) grams.

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Most popular questions from this chapter

Balance the following equations: $$ \begin{array}{l}{\text { (a) } \mathrm{Al}_{4} \mathrm{C}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{CH}_{4}(g)} \\ {\text { (b) } \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (c) } \mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)} \\ {\text { (d) } \mathrm{Mg}_{3} \mathrm{N}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{MgSO}_{4}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q)}\end{array} $$

(a) When a compound containing \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) is completely combusted in air, what reactant besides the hydrocarbon is involved in the reaction? (b) What products form in this reaction? (c) What is the sum of the coefficients in the balanced chemical equation for the combustion of one mole of acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l),\) in air?

A compound, \(\mathrm{KBrO}_{x},\) where \(x\) is unknown, is analyzed and found to contain 52.92\(\%\) Br. What is the value of \(x ?\)

A mixture containing \(\mathrm{KClO}_{3}, \mathrm{K}_{2} \mathrm{CO}_{3}, \mathrm{KHCO}_{3},\) and \(\mathrm{KCl}\) was heated, producing \(\mathrm{CO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) gases according to the following equations: $$ \begin{array}{l}{2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)} \\ {2 \mathrm{KHCO}_{3}(s) \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g)} \\\ {\mathrm{K}_{2} \mathrm{CO}_{3}(s) \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)}\end{array} $$ The KCl does not react under the conditions of the reaction. If 100.0 g of the mixture produces 1.80 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}, 13.20 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and 4.00 \(\mathrm{g}\) of \(\mathrm{O}_{2},\) what was the composition of the original mixture? (Assume complete decomposition of the mixture.)

Calculate the percentage by mass of the indicated element in the following compounds: \((\mathbf{a})\) carbon in acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{a}\) gas used in welding; \((\mathbf{b} )\) hydrogen in ascorbic acid, \(\mathrm{HC}_{6} \mathrm{H}_{7} \mathrm{O}_{6}\) also known as vitamin \(\mathrm{C} ;(\mathbf{c})\) hydrogen in ammonium sulfate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4},\) a substance used as a nitrogen fertilizer; \((\mathbf{d})\) platinum in \(\mathrm{PtCl}_{2}\left(\mathrm{NH}_{3}\right)_{2},\) a chemotherapy agent called cisplatin; \((\mathbf{e})\) oxygen in the female sex hormone estradiol, \(\mathrm{C}_{18} \mathrm{H}_{24} \mathrm{O}_{2} ;(\mathbf{f})\) carbon in capsaicin, \(\mathrm{C}_{18} \mathrm{H}_{27} \mathrm{NO}_{3},\) the compound that gives the hot taste to chili peppers.
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