I need to determine molar weight of CaCO3. Where relative atomic mass of Calcium is 40, Carbon 12 and Oxygen 16.

Sadly i do not understand this kind of physical example yet to solve it by myself. Can you help me?

Guest Mar 14, 2017

#1**0 **

Atomic mass is molar (or 1 mole of an element is this much mass)

CaCO_{3} is the element

Think of it as an addition problem:

Ca + C + O + O + O

Substitute numbers

40+12+16+16+16

Simplify

52+48

Add 52 and 48

100

So, the molar weight (molar mass) of the compound CaCO_{3} is 100 g/mol (grams per mole)

rarinstraw1195
Mar 14, 2017

#2**0 **

rarinstraw1195

Atomic mass is molar (or 1 mole of an element is this much mass)

CaCO3 is the element

Think of it as an addition problem:

Ca + C + O + O + O

Substitute numbers

40+12+16+16+16

Simplify

52+48

Add 52 and 48

100

So, the molar weight (molar mass) of the compound CaCO3 is 100 g/mol (grams per mole)

I love you. Thank you so much. :-)

Guest Mar 14, 2017

#3**0 **

All that molar weight (or molar mass) tells you is how many grams of something are in 1 mole of that thing. That's what atomic masses are on the periodic table- so Calcium weighs 40 grams per mole, Carbon weighs 12 g/mol, etc.

One thing to recognize is that 1 mole of AB = 1 mole A + 1 mole B. Why? Say I had some bags of fruit, and each bag contained 1 apple and 1 orange. If I gave you 100 bags, you would have 100 apples, and 100 oranges, not 50 and 50.

Similarly, If I say I have 1 mol HCl, that means I have 1 mol H + 1 mol Cl:

1 mole H at 1 gram/mole = 1g

1 mole Cl at 35 grams/mole = 35g

In total, 1 mole HCl = 1g + 35g = 36g

So, HCl weighs 36g/mol.

For your example, just follow the same procedure: break it down, find the individual molar masses, and add them back up.

In 1 mole of CaCO_{3}, you have:

1 mole Ca at 40 g/mol = 40g

1 mole C at 12 g/mol = 12g

1 mole O_{3} at (16 x 3) g/mol = 48g (because O_{3} weighs 3x as much as O)

1 mole CaCO_{3} = 40g + 12g + 48g = 100g.

So, the molar mass/weight of CaCO_{3} is 100 grams per mole, or 100 g/mol.

Hope this helped!

Guest Mar 14, 2017