what is the mass of 5.34x10^23 molecules of sodium hydrogen carbonate

Question

i need the calculations

Answer 1

Step 1: Determine the Molar Mass of NaHCO₃

Begin by breaking down sodium hydrogen carbonate into its constituent elements. From there, find the mass of each according to the periodic table. Remember to multiply an element's mass by the subscript (i.e. the number of atoms of that element within the compound). As you will see below, that only pertains to oxygen (x3). Next, add them all together. This sum is the mass of one mol of NaHCO₃.

Na: 23g/mol

H: 1g/mol

C: 12g/mol

O: 16g/mol * 3 = 48 g/mol

+_______________

84g/mol

Step 2: Use dimensional analysis to convert the given "5.34*10²³ molecules of sodium hydrogen carbonate" into grams of sodium hydrogen carbonate (i.e. mass).

Dimensional analysis is a fancy term that simply refers to converting some starting number in some units into the final desired units by using a series of conversion factors. Here, we start with the units of "molecules". We then first use the conversion factor of "1 mol of a substance = 6.02 *10²³ molecules" (also known as Avogadro's number). Notice that when we multiply our starting units of "molecules" by "mols/molecule" in this step, what we ultimately do is allow "molecules" to cancel with "molecules" since anything divided by itself is equal to 1. What we are left with is "mols". Since our desired end result must have the units of "grams" for mass, the next step is to multiply again by another conversion factor. Our conversation factor for this step was found in Step 1 above where we determined that "1 mol of NaHCO₃ = 84 grams". Multiplying "mols" by "grams/mol" allowed "mols" to cancel out, leaving us with the desired units of "grams". All that is left for us to do is to multiply the fractions that we have created to arrive at our final answer.

Answer 2

First get the atomic mass of NaHCO₃ =

Na +H + C + 3*O =

11+1+12+3*16 = 84g/mol

Next convert molecules to moles and multiply by the atomic mass:

5.34*10²³ molecules * 1 mol/6.022*10²³ molecules * 84g/mol =74.49g = 74.5g to 3 significant figures.