Different precision for masses of moon and earth online
We don't know the mass of the Moon with that level of accuracy. NASA gives only 4 significant digits. The best estimate of the mass of the Earth I could find is:
$$M_\oplus = (5.9722 \pm 0.0006)\times 10^{24}\;~{\rm kg}$$
Does this have something to do with the earth mass being a "standard unit of mass in astronomy"?
Yes, most likely. Google must have converted the mass of the Moon in $M_\oplus$ to $\rm kg$ without truncating the digits.
An interesting fact is that we actually know the ratio between the mass of the Moon and the mass of the Earth with an incredible precision:
$$M_☽ = (1.23000371 \pm 0.00000004) × 10^{−2}\;M_\oplus$$
But when you convert this to $\rm kg$, you lose that level of accuracy and end up with:
$$M_☽ = (7.3458 \pm 0.0007) \times 10^{22}\;{\rm kg}$$
One general remark: google does not give a reference for that value. A measured quantity without a source and an error is meaningless. I did not find the mass of the moon in a reputable source with that precision.
For astrophysical constants I can recommend this reference USNO, 2014, Selected Astronomical Constants there the mass of the earth is given by: $$M_E=5.9722(6)\times10^{24}\mathrm{kg}.$$ This value comes from experiments which measure $G M_E$ and then divide by the gravitational constant $G$ to get the mass. The gravitational constant is the limiting factor because it is only known up to $$G=6.67428(15)\times10^{-11}\mathrm{m}^{3}\mathrm{kg}^{-1}\mathrm{s}^{-2}.$$
The mass ratio between earth and moon however can be determined quite well using highly precise position measurements (Lunar Laser Ranging,...) and this methods do not involve $G$:
$$M_M/M_E=1.23000371(4)\times10^{-2}.$$
To my knowledge there are no precise values for the mass of the moon, which would not involve $G$ and using the mass ratio with the mass of the earth will give:
$$M_M=7.3458(7)\times10^{22}\mathrm{kg}.$$
I think the value googles gives has the significant digits of the mass ratio but since the mass of the earth is not know to that precision the digits after the first 4 are meaningless and again a measurement value without a source is worth nothing.
@Wood was faster with his correct answer; but I was nearly finished with this one and it has some additional details.