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

1) To find the number of formula units of calcium carbonate (CaCO3) in a 10.0 gram sample, we need to follow these steps:<br /><br />Step 1: Convert the mass of CaCO3 to moles using its molar mass.<br />Molar mass of CaCO3 = 100.09 g/mol<br />Number of moles = mass / molar mass<br />Number of moles = 10.0 g / 100.09 g/mol<br />Number of moles ≈.100 mol<br /><br />Step 2: Convert the number of moles to the number of formula units using Avogadro's number.<br />Avogadro's number = 6.022 x 10^23 formula units/mol<br />Number of formula units = number of moles x Avogadro's number<br />Number of formula units ≈ 0.100 mol x 6.022 x 10^23 formula units/mol<br />Number of formula units ≈ 6.022 x 10^22 formula units<br /><br />Therefore, there are approximately 6.022 x 10^22 formula units of CaCO3 in a 10.0 gram sample.<br /><br />2) To find the mass of 5.0 x 10^30 molecules of table sugar (C12H22O11), we need to follow these steps:<br /><br />Step 1: Convert the number of molecules to moles using Avogadro's number.<br />Avogadro's number = 6.022 x 10^23 molecules/mol<br />Number of moles = number of molecules / Avogadro's number<br />Number of moles = 5.0 x 10^30 molecules / 6.022 x 10^23 molecules/mol<br />Number of moles ≈ 8.30 x 10^6 mol<br /><br />Step 2: Convert the number of moles to grams using the molar mass of C12H22O11.<br />Molar mass of C12H22O11 = 342.30 g/mol<br />Mass = number of moles x molar mass<br />Mass ≈ 8.30 x 10^6 mol x 342.30 g/mol<br />Mass ≈ 2.84 x 10^9 g<br /><br />Therefore, 5.0 x 10^30 molecules of table sugar (C12H22O11) weigh approximately 2.84 x 10^9 grams.