1) A sample of calcium carbonate (CaCO_(3)) has a mass of 50.0 grams, how many formula units of CaCO_(3) are there? SHOW WORK FOR CREDIT,BOX IN ANSWER 2)How many grams is 25.0times 10^25 molecules of table sugar (C_(12)H_(22)O_(11)) SHOW WORK FOR CREDIT BOX IN ANSWER

1) To find the number of formula units of calcium carbonate (CaCO3) in a 50.0 gram sample, we need to follow these steps:<br /><br />Step 1: Calculate the molar mass of CaCO3.<br />Molar mass of Ca = 40.08 g/mol<br />Molar mass of C = 12.01 g/mol<br />Molar mass of O = 16.00 g/mol<br />Molar mass of CaCO3 = 40.08 + 12.01 + (3 * 16.00) = 100.09 g/mol<br /><br />Step 2: Convert the mass of the sample to moles.<br />Moles of CaCO3 = Mass of sample / Molar mass of CaCO3<br />Moles of CaCO3 = 50.0 g / 100.09 g/mol = 0.500 mol<br /><br />Step 3: Convert moles to formula units using Avogadro's number (6.022 x 10^23 formula units/mol).<br />Number of formula units = Moles of CaCO3 * Avogadro's number<br />Number of formula units = 0.500 mol * (6.022 x 10^23 formula units/mol) = 3.011 x 10^23 formula units<br /><br />Therefore, there are approximately 3.011 x 10^23 formula units of CaCO3 in the 50.0 gram sample.<br /><br />2) To find the mass of 25.0 x 10^25 molecules of table sugar (C12H22O11), we need to follow these steps:<br /><br />Step 1: Calculate the molar mass of C12H22O11.<br />Molar mass of C = 12.01 g/mol<br />Molar mass of H = 1.01 g/mol<br />Molar mass of O = 16.00 g/mol<br />Molar mass of C12H22O11 = (12 * 12.01) + (22 * 1.01) + (11 * 16.00) = 342.30 g/mol<br /><br />Step 2: Convert the number of molecules to moles using Avogadro's number (6.022 x 10^23 molecules/mol).<br />Moles of C12H22O11 = Number of molecules / Avogadro's number<br />Moles of C12H22O11 = (25.0 x 10^25 molecules) / (6.022 x 10^23 molecules/mol) = 4.15 x 10^2 mol<br /><br />Step 3: Convert moles to grams using the molar mass of C12H22O11.<br />Mass of C12H22O11 = Moles of C12H22O11 * Molar mass of C12H22O11<br />Mass of C12H22O11 = (4.15 x 10^2 mol) * (342.30 g/mol) = 1.42 x 10^5 g<br /><br />Therefore, 25.0 x 10^25 molecules of table sugar (C12H22O11) weigh approximately 1.42 x 10^5 grams.