Pagina de inicio
/
Química
/
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

Problemas

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) 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

Solución

avatar
Estelamaestro · Tutor durante 5 años
expert verifiedVerificación de expertos
4.2 (205 votos)

Responder

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 />The molar mass of CaCO3 is the sum of the molar masses of calcium (Ca), carbon (C), and oxygen (O).<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 /><br />Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + (3 * 16.00 g/mol) = 100.09 g/mol<br /><br />Step 2: Convert the mass of the sample to moles.<br />Mass of CaCO3 = 50.0 grams<br />Moles of CaCO3 = Mass / Molar mass<br />Moles of CaCO3 = 50.0 g / 100.09 g/mol = 0.500 mol<br /><br />Step 3: Convert moles to formula units.<br />1 mole of CaCO3 contains Avogadro's number of formula units, which is 6.022 x 10^23 formula units.<br />Formula units of CaCO3 = Moles of CaCO3 * Avogadro's number<br />Formula units of CaCO3 = 0.500 mol * 6.022 x 10^23 formula units/mol<br />Formula units of CaCO3 = 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 />The molar mass of C12H22O11 is the sum of the molar masses of carbon (C), hydrogen (H), and oxygen (O).<br />Molar mass of C = 12.01 g/mol<br />Molar mass of H = 1.008 g/mol<br />Molar mass of O = 16.00 g/mol<br /><br />Molar mass of C12H22O11 = (12 * 12.01 g/mol) + (22 * 1.008 g/mol) + (11 * 16.00 g/mol) = 342.30 g/mol<br /><br />Step 2: Convert the number of molecules to moles.<br />1 mole of C12H22O11 contains Avogadro's number of molecules, which is 6.022 x 10^23 molecules.<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<br />Moles of C12H22O11 = 4.15 x 10^2 mol<br /><br />Step 3: Convert moles to grams.<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<br />Mass of C12H22O11 = 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.
Haz clic para calificar: