Chemestry I

Sorbitol is a sweet substance found in fruits and berries and someties used as a sugar substiture. An aqueous solution containin 1g of sorbitol in 100g water has a freezing point of 0.102C. Elemental analysis indicates that sorbitol is 39.56%C, 7.75%H, and 52.7O. What are the molar mass and molecular formula of sorbitorl? The Kf of water is 1.86C/m

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  1. I answered this question yesterday for someone but I can't find the answer to copy it. Here is what you do; however, let me point out that I think you have a typo in the problem. I'm sure the freezing point is -0.102 and not +0.102.
    There are two problems here.
    A. Problem #1.
    1. Use delta T = Kf*molality. You have delta T and Kf, solve for molality.
    2. Use molality = # mols/kg solvent to calculate mols.
    3. Use mols = g/molar mass to solve for molar mass since you have mols and grams.
    B. Problem #2.
    Then take a 100 g sample which will give you 39.56 g C, 7.75 g H and 52.70 g O.
    Convert those grams to mols, then determine the ratio of C, H, and O in small whole numbers. This will give you the empirical formula.

    Finally, divide the molar mass by the empirical formula mass to determine the number of units of the empirical formula that makes up the molecular formula.

    Post your work if you get stuck.

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  2. It is spelled Chemistry. I would learn to spell that before you go much further.

    Assume 100 grams of the stuff to get a formula.

    then you have 39.56g C, and so on with H and O.

    Figure the moles of C, H, and O from that. Divide by the lowest moles of all the element, in order to get a ratio of C H O. The ratio for the lowest element will be 1, or course. You will get some numbers that will look like this
    C3H4O2 or some such numbers from the ratios. That is the empirical formula. If you don't follow this explaination, go here:
    http://home.c2i.net/astandne/help_htm/english/example/empirical-formula.htm

    Now to get mole mass, use the molatity

    I assume the freezing point is -.102C, not what you listed.
    -.102=-1.86m
    solve for m molality.
    Then, 1g/molmass /.100 = m and you can solve for mole mass.

    Now you have the true molmass, and the empirical formula. Figure the molmass of the empirical formula, divide it into the true mole mass. You should get a whole number, which tells you how much each of the coefficents on the empirical should be increased.

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    bobpursley

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