One method of favoring the formation of an ester is to add excess acetic acid. Suggest another method, involving the right-hand
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<span>When you have 100 g of compound, then based on the percentages given, there are 40.0 g C, 6.70 g H, and 53.3 g O. Convert those to moles:
</span>C: 40.0 g / 12.0 = 3.33 moles of C
<span>H: 6.70 g / 1.01 = 6.63 moles of H </span>
<span>O: 53.3 / 16.0 = 3.33 moles of O
</span>
<span>Dividing by the smallest (3.33), we get a C:H:O mole ratio of 1:2:1
</span>So, <span>The empirical formula is CH2O.
Now, </span><span>That formula has a molar mass of [12.0 + 2(1.0) + 16.0] = 30.0
And we are given it's molar mass is = 240
So, no. of units of CH2O = 240 / 30 = 8
</span><span>8 x CH2O = C8H16O8, and that is the molecular formula.
</span>
</span>C: 40.0 g / 12.0 = 3.33 moles of C
<span>H: 6.70 g / 1.01 = 6.63 moles of H </span>
<span>O: 53.3 / 16.0 = 3.33 moles of O
</span>
<span>Dividing by the smallest (3.33), we get a C:H:O mole ratio of 1:2:1
</span>So, <span>The empirical formula is CH2O.
Now, </span><span>That formula has a molar mass of [12.0 + 2(1.0) + 16.0] = 30.0
And we are given it's molar mass is = 240
So, no. of units of CH2O = 240 / 30 = 8
</span><span>8 x CH2O = C8H16O8, and that is the molecular formula.
</span>