Henessa Gumiran09/25/10Period 09-10

Measurement—the Basis of Quantitative Chemistry

Objectives- To gain skill in making accurate measurements.- To become familiar with the limitations of measuring devices and sources of experimental error.- To obtain a working knowledge of the concept of density.- To determine the density of a rectangular or cylindrical solid.- To determine the thickness of a piece of metal foil.

Data

Data Table I: Direct Measurement of VolumeUnit of Measurement MeasurementMass of object 70.94 gLength of object 5.00 cmWidth or diameter of object 1.25 cmDepth of object (if rectangular) 1.25 cmVolume of object in ml 7.81 mlDensity of object in grams/ml 9.08 g/ml

Data Table II: Volume by Water DisplacementUnit of Measurement MeasurementFinal reading of water level 33.0 mlInitial reading of water level 25.0 mlVolume of water displaced 8.00 mlVolume of object 8.00 mlMass of object 70.94 gDensity of object 8.87 g/ml

Data Table III: Volume by Archimedes’ PrincipleUnit of Measurement MeasurementMass of object in air 70.94 gMass of object in water 63.05 gMass of water displaced by object 7.89 gDensity of water 1.00 g/mlVolume of water displaced by object 7.89 mlVolume of object 7.89 cm3

Density of object 8.99 g/cm3

Data Table V: Thickness of Aluminum FoilUnit of Measurement MeasurementMass 0.43 gLength 11.4 cmWidth 5.4 cmDensity 2.70 g/mlVolume 7.81 mlArea 62. cm2

Thickness (cm) 2.6 x 10-3 cmThickness (Angstroms) 2.6 x 105 ÅThickness (micrometer) 26. µm

Questions01) The densities of several common metals are listed. Compare your experimental density with the given values and try to

determine which metal you tested. Which other property could be used to help identify your sample?The experimental density matches closest with that of copper—8.92 g/ml. Another unique property that could be used to help identify the sample is by testing its melting point.

02) Which method do you think gave the most accurate value for the volume in Part I? Explain your choice.The method that gave the most accurate value for the volume in Part I would be that of the utilization of the Archimedes’ Principle, in that there were minimal measurements made. Thus, there was less chance of error as only the mass of the object in water was taken, as opposed to the multiple measurements in the other methods.

03) List sources of error for each method.With the direct method of measuring volume, the measurements made using the ruler may differ as a result of different readings for measurements between the calibrations that may be set unevenly, and the error would multiply when obtaining the volume. As for the second method using water displacement, the readings of the water levels may have been inaccurate due to an uneven surface, thus leading to a volume that would be off when the amount of water displaced is calculated. Also, for the volume using the Archimedes’ Principle, error may have resulted from the massing of the object in water possibly as a result of things such as the object not being suspended properly.

04) What measurement limited the accuracy in each case?In the direct measurement of volume, the ruler was a source of limitation as measurements could only be read to the nearest millimeter, but the measurement may be in between the calibrations. Thus, the accuracy of the ruler is limited. The graduated cylinder, read to the nearest 0.2 ml, was also a limitation in that the actual readings for the initial and final levels may have been in between. As for the volume by Archimedes’ Principle, the triple beam balance may have limited the accuracy in that the riders may not have led the balance to be zeroed out, and therefore, the measurement of mass would not be accurate.

05) Assuming the balance could be read to 0.01 g, what percentage error might you expect when weighing a 10.00-gram object?The percentage error you might expect when weighing a 10.00g object, you might expect a .1% error.

10.01g−10.00g10.00g

×100=0.1%

06) What percentage error would you expect in using your graduated cylinder to measure 5 ml of water if you can read the graduated cylinder to 0.2 ml?The expected percentage error in using a graduated cylinder read to 0.2 ml when measuring 5 ml of water is 4%.

5.2ml−5.0ml5ml

×100=4%

07) What percentage error would you expect in using a buret to measure 5 ml of water if you can read a buret to 0.02 ml?The percentage error expected in using a buret read to 0.02 ml is 0.4%.

5.02ml−5ml5ml

×100=0.4%

10) A full barrel holds 750 pounds of gasoline. The same barrel holds 1100 pounds of water when full. What is the density of the gasoline?The density of gasoline is .68 g/ml. This is achieved by first calculating the volume of the barrel by using the density of water along with its weight, since it is known that the mass would be equal to the volume. Given the weight of the gasoline, we can find the mass, which would then be divided by the volume of the barrel to find the density of gasoline.

1100 lb ×454 g

lb×1ml1.00 g

=499400ml

750 lb×454 g

lb× ❑499400ml

=.68 g/ml❑

11) A solution of sodium hydroxide consisting of a solute (sodium hydroxide) and a solvent (water) has a density of 1.22 g/ml. What does 0.100 liter of this solution weigh? If the solution is 20.0 percent by mass solute, what mass of pure sodium hydroxide is in the 0.100 liter of solution?0.100 liters of the solution weighs 122 g, after it is converted from liters to grams. If the mass of the solution is 20.0% solute, 24.4 g of pure sodium hydroxide is in 0.100 liter of solution.

.100 L×103mL

L×1.22g

mL=122g

.100 L×103mL

L×1.22g

mL× .200=24.4 g

12) Osmium metal, the densest element, has a density of 22.5 g/ml, while hydrogen gas, the least dense element, has a density of 8.90 x 10-5 g/ml. Calculate the volume occupied by 1.00 g of each element. How many times denser than hydrogen is osmium?To obtain the volume occupied by each, 1.00g is multiplied by the reciprocal of the densities so that the grams cancel. 1.00 grams of osmium metal occupies 4.44 X 10-2 ml, while hydrogen gas, rounded to three significant figures, occupies 1.12 X 104 ml. When the density of osmium is divided by the density of hydrogen, it is found that osmium is 252809 times, or when rounded to three significant figures, 2.53 X 105 times denser than hydrogen.

1.00 g ×mL22.5g

=4.44×10−2g

1.00 g ×mL

8.90×10−5 g=1.12×104 g

22.5gmL

×mL

8.90×10−5g=2.53×105

13) Calculate the approximate thickness of the aluminum foil in terms of atoms. Each aluminum atom has a radius of 1.48 Angstrom units. If each atom has a radius of 1.48 Angstrom units, the diameter per atom would be 2.96 Angstroms, or 2.96 X 10-8 cm per atom. As the calculation for the thickness came to be 2.6 X 10-3 cm, this would be divided by 2.96 X 10-8 cm per atom. Thus, the units cancel, leaving 8.45 x 104 atoms in thickness.

2.6 X10−3 cm2.96 X 10−8 cm /atom

=.878×105=8.78×104atoms