If you burned 4.10 x 10^24 molecules of butane (C4H10), what mass of ethane did you burn?

Explanation:

Molar mass of C4H10 = 12x 4 + 1 x 10= 48+ 10=58g.

Also note that molar mass of a substance is equivalent to Avogadro's number (6.02 X 10^-23)

Hence,

6.02 X 10^-23 molecules of ethane was burned in 58g of ethane

So, 4.10 X 10^24 molecules of ethane will burn in xg of ethane.

Cross multiply:

6.02 X 10^-23x = 58 X 4.10 X 10^24

6.02 X 10^-23x = 2.378 X 10 ^26

Dividing both sides by the coefficient of 'x'

x = 2.378 X 10^26/(6.02X 10^-23)

x = 3.95 X10^48g approximately

x = 4.0 X 10^48g

Or

Mole = number of elementary

particless / Avogadro's number

Where,

No of elementary particles= 4.10 X 10^24 molecules

Avogadro's number= 6.02 x 10^-23

Therefore,

Mole = 4.10 X 10^24/ (6.02 X 10^-23)

= 6.81 X 10^ 46 moles.

But mole = mass / molar mass

Mass = mole x molar mass

where

Molar mass of C4H10 = 12x 4 + 1 x 10= 48+ 10=58g/mol

Mass= 6.81 X 10^46 X 58

= 3.95 X 10^48g approximately

= 4.0 X 10^ 48g.

Thanks

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.