How many C2H4 molecules are contained in 45.8 mg of C2H4? The molar mass of C2H4 is 28.05 g/mol. A) 9.83 \times 10^{20} C2H4 molecules B) 2.71 \times 10^{20} C2H4 molecules C) 7.74 \times 10^{26} C2H4 molecules D) 3.69 \times 10^{23} C2H4 molecules E) 4.69 \times 10^{23} C2H4 molecules
The Correct Answer and Explanation is:
To determine how many C₂H₄ molecules are present in 45.8 mg of C₂H₄, follow these steps:
Step 1: Convert milligrams to grams
Since the molar mass is in grams per mole, convert 45.8 mg to grams:45.8 mg=45.81000=0.0458 g45.8\, \text{mg} = \frac{45.8}{1000} = 0.0458\, \text{g}45.8mg=100045.8=0.0458g
Step 2: Calculate number of moles
Use the molar mass of C₂H₄ (28.05 g/mol) to calculate moles:Moles of C2H4=0.0458 g28.05 g/mol=0.001633 mol\text{Moles of } C_2H_4 = \frac{0.0458\, \text{g}}{28.05\, \text{g/mol}} = 0.001633\, \text{mol}Moles of C2H4=28.05g/mol0.0458g=0.001633mol
Step 3: Convert moles to molecules
Use Avogadro’s number, which is:6.022×1023 molecules/mol6.022 \times 10^{23}\, \text{molecules/mol}6.022×1023molecules/molMolecules of C2H4=0.001633 mol×6.022×1023 molecules/mol\text{Molecules of } C_2H_4 = 0.001633\, \text{mol} \times 6.022 \times 10^{23}\, \text{molecules/mol}Molecules of C2H4=0.001633mol×6.022×1023molecules/mol=9.83×1020 molecules= 9.83 \times 10^{20}\, \text{molecules}=9.83×1020molecules
Answer: A) 9.83×10209.83 \times 10^{20}9.83×1020 C₂H₄ molecules
Explanation
To find the number of ethene (C₂H₄) molecules in a given mass, we first need to understand the relationship between mass, moles, and the number of molecules. The concept of the mole is central in chemistry. A mole represents a specific number of particles, known as Avogadro’s number, which is 6.022×10236.022 \times 10^{23}6.022×1023. The molar mass of a substance tells us how much one mole of that substance weighs. For C₂H₄, the molar mass is 28.05 g/mol, which means one mole of ethene weighs 28.05 grams.
Since the question gives the mass of ethene in milligrams, we start by converting it to grams because molar mass is expressed in grams per mole. After converting 45.8 mg to 0.0458 g, we divide this mass by the molar mass to determine how many moles are present. This gives us approximately 0.001633 moles of C₂H₄.
To find how many molecules this represents, we multiply the number of moles by Avogadro’s number. This multiplication gives us approximately 9.83×10209.83 \times 10^{20}9.83×1020 molecules of ethene. This process shows the importance of unit conversions and the concept of the mole in linking the macroscopic measurement of mass to the microscopic count of individual molecules.
Thus, the correct answer is A.
