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Experiment 6

By Lawrence Kelly,2014-01-29 01:18
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Experiment 6

Chapter 6 Laboratory Name_______________________________

    Nuts and Bolts of Compounds Corey CP Chemistry Period _____________ ( % Composition, Empirical Formula) Date of Lab__________________________

     Partners:

    Purpose: Determine the percent composition of “compounds” made of nuts and bolts. Demonstrate how empirical formulas can be determined from the percent composition.

    Background Information: Atoms of different elements combine in simple whole number ratios to form chemical compounds. The same two elements may form several different compounds by combining in different

    ratios.

Chemical formulas such as HO show by subscripts the number of atoms of each element in a compound. The 2

    simplest whole number ratios by which elements combine are given in what is called the empirical formula of a

    compound. The actual number of atoms of each element in the compound is given in molecular formulas. The

    purpose of this activity is to use a physical model to illustrate the meaning of chemical formulas.

Materials: nuts and bolts (10 per group)

     balance

Procedure:

    1. Mass an individual nut and bolt separately and enter the two masses in the data table below.

    2. Combine a single nut and bolt to form the compound BtNt. Determine the mass of the nut and bolt combination. Enter the mass in your data table.

3. Combine two nuts and one bolt to form the compound Bt Nt. Mass this combination and enter the mass in 2

    the data table.

4. Combine two bolts with one nut to form the compound BtNt. Mass this combination and enter the mass in 2

    the data table.

5. Make any combination of nuts and bolts you wish. Write the formula (BtNt). Mass this combination ______

    and enter the mass in the data table.

Data and Observations:

Mass of bolt g

    Mass of nut g

    Mass of compound BtNt g

    Mass of compound BtNt g 2

    Mass of compound BtNt g 2

    Mass of compound BtNt g _________

Analysis:

    1. Find the percent Nt and Bt in each of your nut and bolt compounds by using the formulas below. Enter the results in the table below.

     mass of bolts mass of nuts

    % bolts (% Bt) = ------------------------------ % nuts (%Bt) = -----------------------------

     total mass of compound total mass of compound

     BtNt BtNt BtNt Bt Nt22

    % Bt

    % Nt

2. For BtNt, carry out the calculations below:

     % bolts % nuts

     mass of one bolt mass of one nut

    Divide the larger of these two calculated numbers by the smaller one. The result is the ratio of bolts to nuts in the empirical formula. Write the empirical formula.

    3. Repeat the calculations for each of your other compounds. Remember that the empirical formula must be the simplest whole number ratio. If your ratios are not whole numbers, they must be converted to whole numbers ( BtNtbecomes BtNt ) or rounded to whole numbers (BtNt becomes BtNt). Write the 0.51.0 20.91.1

    empirical formulas found from your data. Show your work!

Extension and Application

    1. Use your measured masses to calculate the percent composition of BtNt. 22

     Mass of 1 Bolt Mass of 1 Nut

    % Bt = Mass of 2 nuts + 2 bolts x 100 % Nt = Mass of 2 nuts + 2 bolts X 100

    Compare it with the percent composition of BtNt that you calculated (in your Analysis table).

    Assuming you cannot see the combined nuts and bolts, what information could you use to decide whether your invisible compound is BtNt or BtNt? 22

2. Obtain the percent composition of the BtNt compound of another lab group (do NOT find out the ________

    formula!!!). Calculate the empirical formula of their BtNt compound. Now check to see if your ___________

    answer agrees with their compound.

    3. Use your periodic table to determine the empirical formulas for the following compounds:

     a. % H = 6.0% % S = 94.0%

     b. % Na = 21.7% % Cl = 33% % O = 45 %

    4. What is the empirical formula for a compound that has the analysis below?

     2.3 g Na

     1.2 g C

     1.6 g O

     ---------

     5.1 g total sample

Conclusion:

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