A sample of glucose, C6H1206, contains 1.250 x 1021 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? 3.5 2 42 3.41 A sample of glucose, C H12O6, contains 1.250 x 1021 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?
The Correct Answer and Explanation is :
To solve this, let’s break it down step by step:
Known data:
- Chemical formula: ( \text{C}6\text{H}{12}\text{O}_6 ) (Glucose).
- Carbon atoms in the sample: ( 1.250 \times 10^{21} ).
- Atomic composition per molecule: 6 Carbon (C), 12 Hydrogen (H), and 6 Oxygen (O).
(a) Number of hydrogen atoms in the sample:
Each glucose molecule contains 12 hydrogen atoms. The ratio of hydrogen to carbon in a single glucose molecule is:
[
\frac{\text{H atoms}}{\text{C atoms}} = \frac{12}{6} = 2
]
Thus, the number of hydrogen atoms in the sample is:
[
\text{H atoms} = 2 \times \text{C atoms} = 2 \times (1.250 \times 10^{21}) = 2.500 \times 10^{21}
]
(b) Number of glucose molecules in the sample:
Each glucose molecule contains 6 carbon atoms. The number of glucose molecules is:
[
\text{Glucose molecules} = \frac{\text{C atoms}}{\text{C atoms per molecule}} = \frac{1.250 \times 10^{21}}{6}
]
[
\text{Glucose molecules} = 2.083 \times 10^{20}
]
(c) Moles of glucose in the sample:
Avogadro’s number (( N_A )) is ( 6.022 \times 10^{23} ) molecules/mol. Moles of glucose can be calculated as:
[
\text{Moles of glucose} = \frac{\text{Number of molecules}}{N_A} = \frac{2.083 \times 10^{20}}{6.022 \times 10^{23}}
]
[
\text{Moles of glucose} = 3.46 \times 10^{-4} \, \text{mol}
]
(d) Mass of the sample in grams:
The molar mass of glucose (( \text{C}6\text{H}{12}\text{O}_6 )) is:
[
6(\text{C}) + 12(\text{H}) + 6(\text{O}) = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 \, \text{g/mol}
]
Mass of the sample:
[
\text{Mass} = \text{Moles} \times \text{Molar mass} = (3.46 \times 10^{-4}) \times 180.16
]
[
\text{Mass} = 0.0623 \, \text{g}
]
Final answers:
(a) ( 2.500 \times 10^{21} ) H atoms
(b) ( 2.083 \times 10^{20} ) glucose molecules
(c) ( 3.46 \times 10^{-4} \, \text{mol} )
(d) ( 0.0623 \, \text{g} )
Explanation:
The calculation uses the molecular structure of glucose to determine the relationships between the number of carbon, hydrogen, and molecules of glucose. Avogadro’s number connects the molecular count to moles, and the molar mass converts moles to mass. Understanding these fundamental relationships is key to solving such stoichiometric problems systematically.