Problemas
1) A sample of calcium carbonate (CaCO_(3)) has a mass of 100.0 grams, how many formula units of CaCO_(3) are there? SHOW WORK FOR CREDIT BOX IN ANSWER 2) How many grams is 8.0times 10^15 molecules of table sugar (C_(12)H_(22)O_(11)) SHOW WORK FOR CREDIT, BOX IN ANSWER
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Víctor
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Respuesta
1) To find the number of formula units of calcium carbonate (CaCO3) in a 100.0 gram sample, we need to follow these steps:Step 1: Calculate the molar mass of CaCO3.The molar mass of CaCO3 is the sum of the molar masses of calcium (Ca), carbon (C), and oxygen (O).Molar mass of Ca = 40.08 g/molMolar mass of C = 12.01 g/molMolar mass of O = 16.00 g/molMolar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + (3 * 16.00 g/mol) = 100.09 g/molStep 2: Convert the mass of the sample to moles.Mass of CaCO3 = 100.0 gramsMoles of CaCO3 = Mass / Molar massMoles of CaCO3 = 100.0 g / 100.09 g/mol = 0.9995 molesStep 3: Convert moles to formula units.1 mole of CaCO3 contains Avogadro's number of formula units, which is 6.022 x 10^23 formula units.Number of formula units = Moles * Avogadro's numberNumber of formula units = 0.9995 moles * (6.022 x 10^23 formula units/mole) = 5.99 x 10^23 formula unitsTherefore, there are approximately 5.99 x 10^23 formula units of CaCO3 in a 100.0 gram sample.2) To find the mass of 8.0 x 10^15 molecules of table sugar (C12H22O11), we need to follow these steps:Step 1: Calculate the molar mass of C12H22O11.The molar mass of C12H22O11 is the sum of the molar masses of carbon (C), hydrogen (H), and oxygen (O).Molar mass of C = 12.01 g/molMolar mass of H = 1.01 g/molMolar mass of O = 16.00 g/molMolar mass of C12H22O11 = (12 * 12.01 g/mol) + (22 * 1.01 g/mol) + (11 * 16.00 g/mol) = 342.30 g/molStep 2: Convert molecules to moles.1 mole of C12H22O11 contains Avogadro's number of molecules, which is 6.022 x 10^23 molecules.Moles of C12H22O11 = Number of molecules / Avogadro's numberMoles of C12H22O11 = (8.0 x 10^15 molecules) / (6.022 x 10^23 molecules/mole) = 1.33 x 10^-8 molesStep 3: Convert moles to grams.Mass of C12H22O11 = Moles * Molar massMass of C12H22O11 = 1.33 x 10^-8 moles * 342.30 g/mol = 4.56 x 10^-6 gramsTherefore, 8.0 x 10^15 molecules of table sugar (C12H22O11) is equivalent to 4.56 x 10^-6 grams.