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Simora [160]
3 years ago
13

Why can't methanol, CH3OH, be used as a solvent for sodium amide, NaNH2? Sodium amide is nonpolar and methanol is polar. Sodium

amide is polar and methanol is nonpolar. Sodium amide undergoes an acid-base reaction with methanol. There would be no ion-dipole attractive forces between the two compounds.
Chemistry
1 answer:
Elodia [21]3 years ago
4 0

Answer: sodium amide undergoes an acid -base reaction

Explanation:

sodium amide is a ionic compound and basically exists as sodium cation and amide anion. Amide anion is highly basic in nature and hence as soon as  there is amide anion generated in the solution , Due to its very pronounced acidity it very quickly abstracts the slightly acidic proton available on methanol.

This leads to formation of ammonia and sodium methoxide.

Hence sodium amide reacts with methanol and abstracts its only acidic proton and form ammonia and sodium Methoxide.

Hence the 3rd statement is a corrects statement.

So we cannot use methanol for sodium amide because sodium amide itself would react with methanol and the inherent molecular natur of sodium amide would then change.

The 1st and 2nd statements both are incorrect because both the compounds methanol as well as sodium amide have dipole moments and hence are polar molecules.

The 4th statement is also incorrect as both the molecules have dipole moment and hence there would be ion-dipole forces operating between them.

The following reaction occurs:

NaNH₂+CH₃OH→NH₃+CH₃ONa

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A compound has a molecular weight of 146 g/mol. A 0.3250 g sample of the compound contains 0.1605 g of carbon, 0.0220 g of hydro
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Answer: The molecular formula is C_6H_{10}S_2

Explanation:

We are given:

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Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{0.1605g}{12g/mole}=0.0134moles

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.0220g}{1g/mole}=0.0220moles

Moles of S =\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{0.1425g}{32g/mole}=0.0044moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{0.0134}{0.0044}=3

For H = \frac{0.0220}{0.0044}=5

For S =\frac{0.0044}{0.0044}=1

The ratio of C : H: S=  3: 5: 1

Hence the empirical formula is C_3H_5S

The empirical weight of C_3H_5S = 3(12)+5(1)+1(32)= 73g.

The molecular weight = 146 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{146}{73}=2

The molecular formula will be=2\times C_3H_5S=C_6H_{10}S_2

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