Mole Worksheet (Key)

Transcription

1 Mole Worksheet (Key) 1. The average distance between the Earth and Moon is 84,40 km. Using material provided in lab, calculate how many pennies would be needed to stack head to tails to span this distance. *I measured 10 pennies to have a height of 1.40cm. 8440km x 1000m km x 100cm m x 10 pennies 1.40cm = 2.75x1011 pennies 2. Determine the mass of a 5.00 cm line drawn with a number 2 pencil. a. Assuming the pencil lead is made of carbon atoms (diameter = 1.54 x m), calculate the length of these atoms if they were lined end to end. *Determine the mass of blank piece of paper to be grams. Then drew a line on the paper and determined the mass to be grams. Mass of line = (Paper + line) - (Paper with no line) Mass of line = g g = g g C x mole C 12.01gC x 6.022x10 2 C atoms molec x 1.54x10-10 m C atom = x10 9 m = 4x10 9 m b. If the atoms are lined end to end will they reach the moon? If so how many times? x10 9 km m x 1000m trips to moon x =10 trips to moon 8440km

2 . Using material provided in lab, determine the mass (to five significant figures) and volume (to three significant figures) of the average popcorn kernel. *I measured the mass of 18 kernals to be 2.51 grams, and the volume of 66 kernals to be 1.0 ml. a. Calculate the number of popcorn kernels are present in a 1.00 lb bag of popcorn? ( lb = 1 kg) kg 1.00 lb x lb x 1000 g 18 kernals x = 470 kernals kg 2.51g b. I buy popcorn in bulk. I take a 2.00 liter bag with me to the store and fill it 75.0% full. How many popcorn kernels am I purchasing? (0.750)(2.00L) x 1000mL 66 kernals x =7620 kernals L 1.0mL c. The surface area of Texas is km 2. How many miles high would 1.00 moles of popcorn kernels stack on top of Texas? Assume the stack goes straight up and keeps the shape of Texas. (1 mi = km) Volume = (height)(surface area) height = Volume surface area = 1.0mL 66 keranls x 6.022x10 2 kernals x cm mole ml x! m $! km $! mi $ # & # & # & " 100cm % " 1000m % " km %! km 2 mi $ 2 x # & " km % height = mi =106 mi mi

3 4. At the front of the classroom is a vial containing. The empty vial prior to containing the solid sample had a mass of grams. Measure the mass of the vial containing the compound and answer the following questions. a. How many compounds are present in this sample? *I measured the mass of the vial with contents to be grams. Mass of compound = (Mass of vial and contents) - (Mass of empty vial) Mass of compound = ( g) -(11.045g) Mass of compound =.1725g mole(nh.1725g x g x 6.022x10 2 compounds mole x10 21 compounds 4.872x10 21 compounds b. How many nitrogen atoms are in the sample? x N atoms compounds x compounds = 9.74x10 21 N atoms c. How many hydrogen atoms are in the sample? x H atoms 22 compounds x = 9.74x10 H atoms compounds d. What is the mass of water in the sample? mole (NH.1725g x g x 6 moles H 2 mole x g H 2 moles H g H g H 2 e. How many water molecules are in the sample? g H 2 x moles H gh 2 x 6.022x10 2 H 2 molecules = 2.92x10 22 H 2 molecules moles H 2 f. How many hydrogen atoms in this compound are part of water molecules? g H 2 x moles H gh 2 x 6.022x10 2 H 2 molecules moles H 2 2H atoms 22 x = 5.846x10 H atoms H 2 molecules

4 5. Pour approximately 50 ml of water into a 100 ml beaker. Place the beaker on the analytical balance and record the initial mass. Allow the beaker to sit on the balance for two minutes and then record the final mass. Calculate the number of water molecules which evaporated in those two minutes. *I measured the intial mass to be grams. After two minutes the mass was grams. Mass of water that evaporated = grams grams = grams g H 2 x mole H g H 2 x 6.022x10 2 H 2 molecules = 2.x10 20 H 2 molecules mole H 2 6. A sample of sodium metal is determined to have a mass of 5.4 grams. How many moles of sodium are in this sample? 5.4g NaCl x mole NaCl = 0.26 mole NaCl 22.99g NaCl 7. What is the molar mass of C 2 H 6? Molar Mass = 2(12.01g) + 6(1.008g) Molar Mass = 24.02g g Molar Mass = 0.068g Molar Mass = 0.07g 8. What is the mass of moles of C 2 H 6? moles C 2 H 6 x 0.068g C 2 H 6 moles C 2 H 6 = g C 2 H 6 9. A balloon contains 4.29 grams of oxygen. How many oxygen molecules are present in the balloon? 4.29g 2 x mole g 2 x 6.022x10 2 molecules 2 mole 2 = 8.07x10 22 molecules How many oxygen atoms are present in the balloon described in the previous problem? 4.29g 2 x mole 2 x 6.022x10 2 molecules 2 x 2 atoms = 1.61x10 2 atoms 2.00g 2 mole 2 molecule What is the mass of the average helium atom? Answer in grams g He mole He x mole He 6.022x10 2 He atoms = 6.647x10-24 g He/atom He

5 12. How many aluminum atoms are there in a 5.40g sample of aluminum oxide? 5.40g Al 2 x mole Al 2 x 2 moleal x 6.022x10 2 Al atoms g Al 2 mole Al 2 mole Al x10 22 Al atoms 6.414x10 22 Al atoms 1. What mass of aluminum bromide has the same number of aluminum atoms as 5.40g of aluminum oxide? x10 22 Al atoms x 1AlBr compounds 1Al atom mole AlBr x 6.022x10 2 AlBr compounds x g AlBr = 28.40g AlBr mole AlBr 14. In lab you will be reacting a strip of magnesium with oxygen from the air to form magnesium oxide. If the mass of magnesium used is grams, how many magnesium atoms are being used in this reaction? mole Mg 0.057g Mg x 24.1g Mg x 6.022x10 2 Mg atoms = 8.84x10 20 Mg atoms mole Mg 15. What mass of oxygen is needed to react completely with all the magnesium from the previous problem? Hint: what is the ratio of magnesium atoms to oxygen atoms in magnesium oxide? 0.057g Mg x mole Mg 24.1g Mg x mole mole Mg x 16.00g mole = 0.025g 16. What mass of chlorine contains 1.96 x chlorine atoms? 1.96x10 21 mole Cl Cl atoms x 6.022x10 2 Cl atoms x mole Cl 2 2 mole Cl x 70.90g Cl 2 = g Cl 2 mole Cl How many electrons are there in 2.45 grams of calcium? mole Ca 2.45g Ca x 40.08g Ca x 6.022x10 2 Ca atoms x mole Ca 20 electrons Ca atom = 7.6x10 2 electrons 18. The distance (based on a direct path) between San Francisco and Los Angeles is km. Calculate how many iodine atoms (diameter = 2.66 x m) would be needed to stack end to end to span this distance km x 1000m km x I atom 2.66x10-10 m = 2.10x1015 I atoms

6 19. A compound is composed of lead and oxygen. The mass of a sample of this compound is g. The sample contains grams of lead. a. How much oxygen is in this sample? Mass of oxygen = Sample Mass - Mass of lead Mass of oxygen = g g Mass of oxygen = 0.171g b. How many moles of oxygen atoms are in this sample? 0.171g x mole 16.00g = 8.569x10 - moles c. How many moles of lead atoms are in this sample? g Pb x mole Pb 207.2g Pb = 4.28x10 - moles Pb d. What is the chemical formula of this sample? x10 /Pb Molar Ratio = 4.28x10 = 2.000, there are 2 oxygen atoms for every 1 lead atom, (Pb ) - 2 e. What is the chemical name of this sample? Lead (IV) oxide