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Talja [164]
2 years ago
7

What amount (moles) of compound is present in 1.00 g of each of the compounds in Exercise 54?

Chemistry
1 answer:
Anika [276]2 years ago
5 0

Answer:

1. Caffeine, C₈H₁₀N₄O₂

Amount = 1.00/194 = 0.00515 moles

2. Ethanol, C₂H₅OH

Amount = 0.0217 moles

3. Dry Ice, CO₂

amount = 0.0227 moles

<em>Note: The question is incomplete. The compound are as follows:</em>

<em> 1. Caffeine, C₈H₁₀N₄O₂;</em>

<em>2. Ethanol, C₂H₅OH;</em>

<em>3. Dry Ice, CO₂</em>

Explanation:

Amount (moles) = mass in grams /molar mass in grams per mole

1. Caffeine, C₈H₁₀N₄O₂

molar mass of caffeine = 194 g/mol

Amount = 1.00 g/194 g/mol = 0.00515 moles

2. Ethanol, C₂H₅OH

molar mass of ethanol = 46 g/mol

Amount = 1.00 g/46 g/mol = 0.0217 moles

3. Dry Ice, CO₂

molar mass of dry ice = 44 g/mol

amount = 1.00 g/44 g/mol = 0.0227 moles

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