12. Find the mass in grams, of 1.00times 10^23 molecules of N_(2) 13. How many particles are there in 1.43 g of a compound with a gram formula mass of 233 g? 14. How many grams are there in 3.4times 10^24 molecules of NH_(3) 15. Aspartame is an artificial sweetener therts 760 times speeter than sucrose (table sugar)when dissolved in water It is market to by G.D. Searle as Nutra Sweet.The molecular formula of aspartame is C_(14)H_(1)O_(2)O_(5) a) Calculate the gram formula-mass of aspartame. b) How many molecules are in 10 g of aspartame?

12. To find the mass of $1.00\times 10^{23}$ molecules of $N_{2}$, we need to use Avogadro's number, which is $6.022\times 10^{23}$ molecules/mol. <br /><br />First, we need to find the molar mass of $N_{2}$. The atomic mass of nitrogen (N) is approximately 14.01 g/mol. Since $N_{2}$ has two nitrogen atoms, its molar mass is $2 \times 14.01 = 28.02$ g/mol.<br /><br />Now, we can calculate the mass of $1.00\times 10^{23}$ molecules of $N_{2}$:<br /><br />Mass = (Number of molecules) x (Molar mass) / (Avogadro's number)<br />Mass = ($1.00\times 10^{23}$ molecules) x (28.02 g/mol) / ($6.022\times 10^{23}$ molecules/mol)<br />Mass ≈ 4.65 g<br /><br />Therefore, the mass of $1.00\times 10^{23}$ molecules of $N_{2}$ is approximately 4.65 grams.<br /><br />13. To find the number of particles in 1.43 g of a compound with a gram formula mass of 233 g, we can use the concept of moles.<br /><br />First, we need to calculate the number of moles in 1.43 g of the compound:<br /><br />Number of moles = Mass / Molar mass<br />Number of moles = 1.43 g / 233 g/mol<br />Number of moles ≈ 0.00612 mol<br /><br />Now, we can calculate the number of particles using Avogadro's number:<br /><br />Number of particles = Number of moles x Avogadro's number<br />Number of particles = 0.00612 mol x ($6.022\times 10^{23}$ particles/mol)<br />Number of particles ≈ $3.69\times 10^{21}$ particles<br /><br />Therefore, there are approximately $3.69\times 10^{21}$ particles in 1.43 g of the compound.<br /><br />14. To find the mass of $3.4\times 10^{24}$ molecules of $NH_{3}$, we need to use Avogadro's number and the molar mass of $NH_{3}$.<br /><br />The molar mass of $NH_{3}$ is calculated as follows:<br />Molar mass of N = 14.01 g/mol<br />Molar mass of H = 1.01 g/mol<br />Molar mass of $NH_{3}$ = 14.01 g/mol + (3 x 1.01 g/mol) = 17.04 g/mol<br /><br />Now, we can calculate the mass of $3.4\times 10^{24}$ molecules of $NH_{3}$:<br /><br />Mass = (Number of molecules) x (Molar mass) / (Avogadro's number)<br />Mass = ($3.4\times 10^{24}$ molecules) x (17.04 g/mol) / ($6.022\times 10^{23}$ molecules/mol)<br />Mass ≈ 95.5 g<br /><br />Therefore, the mass of $3.4\times 10^{24}$ molecules of $NH_{3}$ is approximately 95.5 grams.<br /><br />15. a) To calculate the gram formula-mass of aspartame ($C_{14}H_{18}N_{2}O_{5}$), we need to sum the atomic masses of all the atoms in the molecule:<br /><br />Atomic mass of C = 12.01 g/mol<br />Atomic mass of H = 1.01 g/mol<br />Atomic mass of N = 14.01 g/mol<br />Atomic mass of O = 16.00 g/mol<br /><br />Gram formula-mass of aspartame = (14 x 12.01) + (18 x 1.01) + (2 x 14.01) + (5 x 16.00)<br />Gram formula-mass of aspartame = 168.14 + 18.18 + 28.02 + 80.00<br />Gram formula-mass of aspartame = 294.34 g/mol<br /><br />Therefore, the gram formula-mass of aspartame is 294.34 g/mol.<br /><br />b) To find the number of molecules in 10 g of aspartame, we can use the concept of moles.<br /><br />First, we need to calculate the number of moles in 10 g of aspartame:<br /><br />Number of moles = Mass / Molar mass<br />Number of moles = 10 g / 294.34 g/mol<br />Number of moles ≈ 0.034 mol<br /><br />Now, we can calculate the number of molecules using Avogadro's number:<br /><br />Number of molecules = Number of moles x Avogadro's number<br />Number of molecules = 0.034 mol x ($6.022\times