Problemas
Chemistry Quiz: Two-Step Mole Conversions 1) A sample of calcium carbonate (CaCO_(3)) has a mass of 10.0 grams, how many formula units of CaCO_(3) are there? SHOW WORK FOR CREDIT BOX IN ANSWER 2) How many grams is 5.0times 10^30 molecules of table sugar (C_(12)H_(22)O_(11)) SHOW WORK FOR CREDIT BOXIN ANSWER
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José
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1) To find the number of formula units of calcium carbonate (CaCO3) in a 10.0 gram sample, we need to follow these steps:Step 1: Convert the mass of CaCO3 to moles using its molar mass.Molar mass of CaCO3 = 100.09 g/molNumber of moles = mass / molar massNumber of moles = 10.0 g / 100.09 g/molNumber of moles ≈.100 molStep 2: Convert the number of moles to the number of formula units using Avogadro's number.Avogadro's number = 6.022 x 10^23 formula units/molNumber of formula units = number of moles x Avogadro's numberNumber of formula units ≈ 0.100 mol x 6.022 x 10^23 formula units/molNumber of formula units ≈ 6.022 x 10^22 formula unitsTherefore, there are approximately 6.022 x 10^22 formula units of CaCO3 in a 10.0 gram sample.2) To find the mass of 5.0 x 10^30 molecules of table sugar (C12H22O11), we need to follow these steps:Step 1: Convert the number of molecules to moles using Avogadro's number.Avogadro's number = 6.022 x 10^23 molecules/molNumber of moles = number of molecules / Avogadro's numberNumber of moles = 5.0 x 10^30 molecules / 6.022 x 10^23 molecules/molNumber of moles ≈ 8.30 x 10^6 molStep 2: Convert the number of moles to grams using the molar mass of C12H22O11.Molar mass of C12H22O11 = 342.30 g/molMass = number of moles x molar massMass ≈ 8.30 x 10^6 mol x 342.30 g/molMass ≈ 2.84 x 10^9 gTherefore, 5.0 x 10^30 molecules of table sugar (C12H22O11) weigh approximately 2.84 x 10^9 grams.