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 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
Roztwór
Martín
avanzado · Tutor durante 1 años
4
(179 Votos)
Respuesta
1) To find the number of formula units of calcium carbonate (CaCO3) in a 100.0 gram sample, we need to use Avogadro's number, which is 6.022 x 10^23 formula units per mole.First, we need to calculate the molar mass of CaCO3:Ca: 40.08 g/molC: 12.01 g/molO: 16.00 g/mol x 3 = 48.00 g/molTotal molar mass of CaCO3 = 40.08 + 12.01 + 48.00 = 100.09 g/molNext, we can calculate the number of moles of CaCO3 in the 100.0 gram sample:Number of moles = mass / molar massNumber of moles = 100.0 g / 100.09 g/mol = 0.9995 molesFinally, we can calculate the number of formula units of CaCO3:Number of formula units = Number of moles x Avogadro's numberNumber of formula units = 0.9995 moles x 6.022 x 10^23 formula units/mol = 5.99 x 10^23 formula unitsTherefore, there are approximately 5.99 x 10^23 formula units of CaCO3 in the 100.0 gram sample.2) To find the mass of 8.0 x 10^15 molecules of table sugar (C12H22O11), we need to use Avogadro's number and the molar mass of C12H22O11.First, we need to calculate the molar mass of C12H22O11:C: 12.01 g/mol x 12 = 144.12 g/molH: 1.01 g/mol x 22 = 22.22 g/molO: 16.00 g/mol x 11 = 176.00 g/molTotal molar mass of C12H22O11 = 144.12 + 22.22 + 176.00 = 342.34 g/molNext, we can calculate the number of moles of C12H22O11 in 8.0 x 10^15 molecules:Number of moles = Number of molecules / Avogadro's numberNumber of moles = 8.0 x 10^15 molecules / 6.022 x 10^23 molecules/mol = 1.33 x 10^-8 molesFinally, we can calculate the mass of C12H22O11:Mass = Number of moles x Molar massMass = 1.33 x 10^-8 moles x 342.34 g/mol = 4.56 x 10^-6 gramsTherefore, 8.0 x 10^15 molecules of table sugar (C12H22O11) have a mass of approximately 4.56 x 10^-6 grams.