An unknown compound containing only C and H was burnt, yielding 10.2 g of CO2 and 6.3 g of H2O. With a molecular weight of about
1 answer:
Answer:
Explanation:
Hello.
In this case, we can see that the mass of carbon of the unknown compound comes from the yielded mass of carbon dioxide, thus, we compute the moles of carbon as follows:
Moreover, the mass of hydrogen comes from the yielded water, therefore we can also compute the moles of water:
Then, to find the subscripts in the empirical formula, we divide by the moles of carbon as the smallest:
Whose molar mass is:
Thus, the ratio of the molecular formula to the empirical formula is:
Therefore, the molecular formula is twice the empirical formula:
Which is actually ethane.
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Answer:
The answer is E. All of the statements describe the anomeric carbon.
Explanation:
When a sugar switches from its open form to its ring form, the carbon from the carbonyl (aldehyde if it is an aldose, or a ketone in the case of a ketose) suffers a nucleophilic addition by one of the hydroxyls in the chain, preferably one that will form a 5 or 6 membered ring after the reaction.
As such, the anomeric carbon will have two oxygens attached (The original one and the one that bonded when the ring closed).
It will be chiral, given that it has 4 different groups attached. (-OR,-OH,-H and -R, where R is the carbon chain).
The hydroxyl group can be in any position (Above of below the ring), depending on with side the addition took place. (See attachment)
It is the carbon of the carbonyl in the open-chain form of the sugar, because it is the only one that can react with the Hydroxyls.
