Calculate each of the following in 0.185 mol of dipropyl ether, (c3h7)2o:
moles of c
moles of o
atoms of h
atoms of c

We have 6 moles of C (2*3=6), 14 moles of H (2*7=14) and 1 mol of O in 1 mol of (C₃H₇)₂O:

Therefore:
1 mol of (C₃H₇)₂O6 moles of C
0.185 moles of (C₃H₇)₂O    x

x=(0.185 moles * 6 moles)/1 mole=1.11 moles

there are 1.11 moles of C in 0.185 moles of dipropyl ether

1 mol of C6.022*10²³ atoms
1.11 moles of C  x

x=(1.11 moles of C *6.022*10²³ atoms)/1 mole of C=6.68 *10²³ atoms

there are 6.68*10²³ atoms of C in 0.185 moles of dipropyl ether.


1 mole of (C₃H₇)₂O14 moles of H
0.185 moles of (C₃H₇)₂O    x

x=(0.185 moles * 14 moles)/1 mole=2.59 moles


 there are 2.59 moles of H in 0.185 moles of dipropyl ether

1 mol 6.022*10²³ atoms
2.59 moles    x

x=(2.59 moles * 6.022*10²³ atoms)/1 mol=1.56*10²⁴ atoms

there are 1.56*10²⁴ atoms in 0.185 moles of dipropyl ether.


1 mol of (C₃H₇)₂O1 moles of C
0.185 moles of (C₃H₇)    x

x=(0.185 moles *1mol)/1 mol=0.185 moles

there are 0.185 moles of O in 0.185 moles of dipropyl ether.

Spontaneity tell us the direction of the reaction, but not how fast it goes. gibbs free energy brings together enthalpy, entropy and temperature in one easy formula. g = h - ts. and, finally the sign of delta g is always negative for a spontaneous reaction.